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Ellipse — JEE Previous Year Questions

Every Ellipse question asked in JEE Main and JEE Advanced across the last 186 papers — 114 questions, each with its correct answer. Free to read, no account needed.

Questions

114

Papers it appeared in

103/186

Appearance rate

55%

All 114 Ellipse questions

Most recent papers first.

Q1·MathematicsNumericalJEE Advanced 2026
Passage: Consider the ellipses given by x2+4y2=1x^{2} + 4y^{2} = 1x2+4y2=1 and 4x2+y2=14x^{2} + y^{2} = 14x2+y2=1. Question: Let PPP be the point in the first quadrant where the given ellipses intersect. If θ\thetaθ is the acute angle between the tangents to the given ellipses at the point PPP, then the value of 4tan⁡θ4\tan\theta4tanθ is _____.

Correct answer: 7.5

Step-by-step solution →
Q2·MathematicsSingle correctJEE Main 2026
Let x2f(a2+7a+3)+y2f(3a+15)=1\dfrac{x^2}{f(a^2 + 7a + 3)} + \dfrac{y^2}{f(3a + 15)} = 1f(a2+7a+3)x2​+f(3a+15)y2​=1 represent an ellipse with major axis along yyy-axis, where fff is a strictly decreasing positive function on R\mathbb{R}R. If the set of all possible values of aaa is R−[α,β]\mathbb{R} - [\alpha, \beta]R−[α,β], then α2+β2\alpha^2 + \beta^2α2+β2 is equal to:
  1. (A)282828
  2. (B)404040
  3. (C)616161
  4. (D)242424

Correct answer: (B)

Step-by-step solution →
Q3·MathematicsSingle correctJEE Main 2026
Let x=9x = 9x=9 be a directrix of an ellipse EEE, whose centre is at the origin and eccentricity is 13\frac{1}{3}31​. Let P(α,0)P(\alpha, 0)P(α,0), α>0\alpha > 0α>0, be a focus of EEE and ABABAB be a chord passing through PPP. Then the locus of the mid point of ABABAB is :
  1. (A)9y2=8x(1−x)9y^{2} = 8x(1 - x)9y2=8x(1−x)
  2. (B)3y2=4x(1−x)3y^{2} = 4x(1 - x)3y2=4x(1−x)
  3. (C)9y2=8x(x−1)9y^{2} = 8x(x - 1)9y2=8x(x−1)
  4. (D)3y2=4x(x−1)3y^{2} = 4x(x - 1)3y2=4x(x−1)

Correct answer: (A)

Step-by-step solution →
Q4·MathematicsSingle correctJEE Main 2026
Let a focus of the ellipse E:x2a2+y2b2=1E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1E:a2x2​+b2y2​=1 be S(4,0)S(4,0)S(4,0) and its eccentricity be 45\frac{4}{5}54​. If the point P(3,α)P(3, \alpha)P(3,α) lies on EEE and OOO is the origin, then the area of △POS\triangle POS△POS is equal to:
  1. (A)12/5
  2. (B)14/5
  3. (C)24/5
  4. (D)48/5

Correct answer: (C)

Step-by-step solution →
Q5·MathematicsNumericalJEE Main 2026
Consider the parabola P:y2=4kxP : y^2 = 4kxP:y2=4kx and the ellipse E:x2a2+y2b2=1E : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1E:a2x2​+b2y2​=1. Let the line segment joining the points of intersection of PPP and EEE, be their latus rectums. If the eccentricity of EEE is eee, then e2+22e^2 + 2\sqrt{2}e2+22​ is equal to _______.

Correct answer: 3

Step-by-step solution →
Q6·MathematicsSingle correctJEE Main 2026
Let an ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1a2x2​+b2y2​=1, a<ba<ba<b, pass through the point (4,3)(4,3)(4,3) and have eccentricity 53\frac{\sqrt{5}}{3}35​​. Then the length of its latus rectum is :
  1. (A)453\frac{4\sqrt{5}}{3}345​​
  2. (B)252\sqrt{5}25​
  3. (C)753\frac{7\sqrt{5}}{3}375​​
  4. (D)853\frac{8\sqrt{5}}{3}385​​

Correct answer: (D)

Step-by-step solution →
Q7·MathematicsNumericalJEE Main 2026
Let A be the point (3,0)(3, 0)(3,0) and circles with variable diameter AB touch the circle x2+y2=36x^2 + y^2 = 36x2+y2=36 internally. Let the curve C be the locus of the point B. If the eccentricity of C is eee, then 72e272e^272e2 is equal to ________.

Correct answer: 18

Step-by-step solution →
Q8·MathematicsSingle correctJEE Main 2026
An ellipse has its center at (1,-2), one focus at (3,-2) and one vertex at (5, - 2). Then the length of its latus rectum is :
  1. (A)163\dfrac{16}{\sqrt{3}}3​16​
  2. (B)6
  3. (C)434\sqrt{3}43​
  4. (D)636\sqrt{3}63​

Correct answer: (B)

Step-by-step solution →
Q9·MathematicsSingle correctJEE Main 2026
Let each of the two ellipses E1:x2a2+y2b2=1,(a>b)E_{1}:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,(a>b)E1​:a2x2​+b2y2​=1,(a>b) and E2:x2A2+y2B2=1E_{2}:\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1E2​:A2x2​+B2y2​=1 , (A<B)(A<B)(A<B) have eccentricity 45\frac{4}{5}54​ . Let the lengths of the latus recta of E1E_{1}E1​ and E2E_{2}E2​ be ℓ1\ell_{1}ℓ1​ and ℓ2\ell_{2}ℓ2​, respectively, such that 2ℓ12=9ℓ22\ell_{1}^{2}=9\ell_{2}2ℓ12​=9ℓ2​ . If the distance between the foci of E1E_{1}E1​ is 8, then the distance between the foci of E2E_{2}E2​ is
  1. (A)965\frac{96}{5}596​
  2. (B)325\frac{32}{5}532​
  3. (C)165\frac{16}{5}516​
  4. (D)85\frac{8}{5}58​

Correct answer: (B)

Step-by-step solution →
Q10·MathematicsNumericalJEE Main 2026
Let (h, k) lie on the circle C:x2+y2=4C : x^{2} + y^{2} = 4C:x2+y2=4 and the point (2h+1,3k+2)(2h + 1, 3k + 2)(2h+1,3k+2) lie on an ellipse with eccentricity e. Then the value of 5e2\dfrac{5}{e^{2}}e25​ is equal to ________.

Correct answer: 9

Step-by-step solution →
Q11·MathematicsSingle correctJEE Main 2026
Let the length of the latus rectum of an ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1a2x2​+b2y2​=1, (a>b)(a > b)(a>b), be 30. If its eccentricity is the maximum value of the function f(t)=−34+2t−t2f(t) = -\frac{3}{4} + 2t - t^{2}f(t)=−43​+2t−t2, then (a2+b2)(a^{2} + b^{2})(a2+b2) is equal to -
  1. (A)516
  2. (B)256
  3. (C)496
  4. (D)276

Correct answer: (C)

Step-by-step solution →
Q12·MathematicsSingle correctJEE Main 2026
Let the line y - x = 1 intersect the ellipse x22+y21=1\frac{x^2}{2} + \frac{y^2}{1} = 12x2​+1y2​=1 at the points A and B. Then the angle made by the line segment AB at the center of the ellipse is :
  1. (A)π−tan⁡−1(14)\pi - \tan^{-1}\left(\frac{1}{4}\right)π−tan−1(41​)
  2. (B)π2+tan⁡−1(14)\frac{\pi}{2} + \tan^{-1}\left(\frac{1}{4}\right)2π​+tan−1(41​)
  3. (C)π2+2tan⁡−1(14)\frac{\pi}{2} + 2\tan^{-1}\left(\frac{1}{4}\right)2π​+2tan−1(41​)
  4. (D)π2−tan⁡−1(14)\frac{\pi}{2} - \tan^{-1}\left(\frac{1}{4}\right)2π​−tan−1(41​)

Correct answer: (B)

Step-by-step solution →
Q13·MathematicsSingle correctJEE Main 2026
Let S and S' be the foci of the ellipse x225+y29=1\frac{x^2}{25} + \frac{y^2}{9} = 125x2​+9y2​=1 and P (α, β) be a point on the ellipse in the first quadrant. If (SP)2+(S′P)2−SP∙S′P=37(SP)^2 + (S'P)^2 - SP \bullet S'P = 37(SP)2+(S′P)2−SP∙S′P=37, then α2+β2α^2 + β^2α2+β2 is equal to :
  1. (A)15
  2. (B)11
  3. (C)17
  4. (D)13

Correct answer: (D)

Step-by-step solution →
Q14·MathematicsSingle correctJEE Main 2026
In the line αx+4y=7\alpha x + 4y = \sqrt{7}αx+4y=7​, where α ∈ R, touches the ellipse 3x2+4y2=13x^2 + 4y^2 = 13x2+4y2=1 at the point P in the first quadrant, then one of the focal distances of P is :
  1. (A)13−1211\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{11}}3​1​−211​1​
  2. (B)13+125\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{5}}3​1​+25​1​
  3. (C)13−125\frac{1}{\sqrt{3}} - \frac{1}{2\sqrt{5}}3​1​−25​1​
  4. (D)13+127\frac{1}{\sqrt{3}} + \frac{1}{2\sqrt{7}}3​1​+27​1​

Correct answer: (D)

Step-by-step solution →
Q15·MathematicsMultiple correctJEE Advanced 2025
Let P(x1,y1)P(x_1, y_1)P(x1​,y1​) and Q(x2,y2)Q(x_2, y_2)Q(x2​,y2​) be two distinct points on the ellipse x29+y24=1\frac{x^2}{9} + \frac{y^2}{4} = 19x2​+4y2​=1 such that y1>0y_1 > 0y1​>0 and y2>0y_2 > 0y2​>0. Let C denote the circle x2+y2=9x^2 + y^2 = 9x2+y2=9, and M be the point (3, 0). Suppose the line x=x1x = x_1x=x1​ intersects C at R, and the line x=x2x = x_2x=x2​ intersects C at S, such that the y-coordinates of R and S are positive. Let ∠ROM=π6\angle ROM = \frac{\pi}{6}∠ROM=6π​ and ∠SOM=π3\angle SOM = \frac{\pi}{3}∠SOM=3π​, where O denotes the origin (0, 0). Let |XY| denote the length of the line segment XY. Then which of the following statements is (are) TRUE?
  1. (A)The equation of the line joining P and Q is 2x+3y=3(1+3)2x + 3y = 3\left(1 + \sqrt{3}\right)2x+3y=3(1+3​)
  2. (B)The equation of the line joining P and Q is 2x+y=3(1+3)2x + y = 3\left(1 + \sqrt{3}\right)2x+y=3(1+3​)
  3. (C)If N2=(x2,0)N_2 = (x_2, 0)N2​=(x2​,0), then 3∣N2Q∣=2∣N2S∣3|N_2Q| = 2|N_2S|3∣N2​Q∣=2∣N2​S∣
  4. (D)If N1=(x1,0)N_1 = (x_1, 0)N1​=(x1​,0), then 9∣N1P∣=4∣N1R∣9|N_1P| = 4|N_1R|9∣N1​P∣=4∣N1​R∣

Correct answer: (A), (C)

Step-by-step solution →
Q16·MathematicsSingle correctJEE Main 2025
Let e1e_1e1​ and e2e_2e2​ be the eccentricities of the ellipse x2b2+y225=1\dfrac{x^2}{b^2}+\dfrac{y^2}{25}=1b2x2​+25y2​=1 and the hyperbola x216−y2b2=1\dfrac{x^2}{16}-\dfrac{y^2}{b^2}=116x2​−b2y2​=1, respectively. If b<5b<5b<5 and e1e2=1e_1e_2=1e1​e2​=1, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:
  1. (A)45\dfrac{4}{5}54​
  2. (B)35\dfrac{3}{5}53​
  3. (C)74\dfrac{\sqrt{7}}{4}47​​
  4. (D)32\dfrac{\sqrt{3}}{2}23​​

Correct answer: (B)

Step-by-step solution →
Q17·MathematicsSingle correctJEE Main 2025
Let the length of a latus rectum of an ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1a2x2​+b2y2​=1 be 10. If its eccentricity is the minimum value of the function f(t)=t2+t+1112f(t)=t^2+t+\dfrac{11}{12}f(t)=t2+t+1211​, t∈Rt\in\mathbb{R}t∈R, then a2+b2a^2+b^2a2+b2 is equal to:
  1. (A)125
  2. (B)126
  3. (C)120
  4. (D)115

Correct answer: (B)

Step-by-step solution →
Q18·MathematicsSingle correctJEE Main 2025
The length of the latus-rectum of the ellipse, whose foci are (2,5)(2,5)(2,5) and (2,−3)(2,-3)(2,−3) and eccentricity is 45\dfrac{4}{5}54​, is
  1. (A)65\dfrac{6}{5}56​
  2. (B)503\dfrac{50}{3}350​
  3. (C)103\dfrac{10}{3}310​
  4. (D)185\dfrac{18}{5}518​

Correct answer: (D)

Step-by-step solution →
Q19·MathematicsSingle correctJEE Main 2025
Let for two distinct values of ppp the lines y=x+py=x+py=x+p touch the ellipse E:x242+y232=1E:\dfrac{x^2}{4^2}+\dfrac{y^2}{3^2}=1E:42x2​+32y2​=1 at the points AAA and BBB. Let the line y=xy=xy=x intersect EEE at the points CCC and DDD. Then the area of the quadrilateral ABCDABCDABCD is equal to:
  1. (A)36
  2. (B)24
  3. (C)48
  4. (D)20

Correct answer: (B)

Step-by-step solution →
Q20·MathematicsSingle correctJEE Main 2025
The centre of a circle CCC is at the centre of the ellipse E:x2a2+y2b2=1E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1E:a2x2​+b2y2​=1, a>ba>ba>b. Let CCC pass through the foci F1F_1F1​ and F2F_2F2​ of EEE such that the circle CCC and the ellipse EEE intersect at four points. Let PPP be one of these four points. If the area of the triangle PF1F2PF_1F_2PF1​F2​ is 30 and the length of the major axis of EEE is 17, then the distance between the foci of EEE is:
  1. (A)26
  2. (B)13
  3. (C)12
  4. (D)132\dfrac{13}{2}213​

Correct answer: (B)

Step-by-step solution →
Q21·MathematicsSingle correctJEE Main 2025
Let CCC be the circle of minimum area enclosing the ellipse E:x2a2+y2b2=1E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1E:a2x2​+b2y2​=1 with eccentricity 12\dfrac{1}{2}21​ and foci (±2,0)(\pm 2,0)(±2,0). Let PQRPQRPQR be a variable triangle, whose vertex PPP is on the circle CCC and the side QRQRQR of length 2a2a2a is parallel to the major axis of EEE and contains the point of intersection of EEE with the negative yyy-axis. Then the maximum area of the triangle PQRPQRPQR is:
  1. (A)6(3+2)6(3+\sqrt2)6(3+2​)
  2. (B)8(3+2)8(3+\sqrt2)8(3+2​)
  3. (C)6(2+3)6(2+\sqrt3)6(2+3​)
  4. (D)8(2+3)8(2+\sqrt3)8(2+3​)

Correct answer: (D)

Step-by-step solution →
Q22·MathematicsSingle correctJEE Main 2025
A line passing through the point P(5,5)P(\sqrt5,\sqrt5)P(5​,5​) intersects the ellipse x236+y225=1\dfrac{x^2}{36}+\dfrac{y^2}{25}=136x2​+25y2​=1 at AAA and BBB such that (PA)⋅(PB)(PA)\cdot(PB)(PA)⋅(PB) is maximum. Then 5(PA2+PB2)5(PA^2+PB^2)5(PA2+PB2) is equal to:
  1. (A)218
  2. (B)377
  3. (C)290
  4. (D)338

Correct answer: (D)

Step-by-step solution →
Q23·MathematicsSingle correctJEE Main 2025
If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is:
  1. (A)417\dfrac{4}{\sqrt{17}}17​4​
  2. (B)316\dfrac{\sqrt3}{16}163​​
  3. (C)319\dfrac{3}{\sqrt{19}}19​3​
  4. (D)57\dfrac{\sqrt5}{7}75​​

Correct answer: (A)

Step-by-step solution →
Q24·MathematicsSingle correctJEE Main 2025
If SSS and S′S'S′ are the foci of the ellipse x218+y29=1\dfrac{x^2}{18}+\dfrac{y^2}{9}=118x2​+9y2​=1 and PPP be a point on the ellipse, then min⁡(SP⋅S′P)+max⁡(SP⋅S′P)\min(SP\cdot S'P)+\max(SP\cdot S'P)min(SP⋅S′P)+max(SP⋅S′P) is equal to:
  1. (A)3(1+2)3(1+\sqrt{2})3(1+2​)
  2. (B)3(6+2)3(6+\sqrt{2})3(6+2​)
  3. (C)9
  4. (D)27

Correct answer: (D)

Step-by-step solution →
Q25·MathematicsSingle correctJEE Main 2025
If αx+βy=109\alpha x+\beta y=109αx+βy=109 is the equation of the chord of the ellipse x29+y24=1\dfrac{x^2}{9}+\dfrac{y^2}{4}=19x2​+4y2​=1, whose mid point is (52,12)\left(\dfrac{5}{2},\dfrac{1}{2}\right)(25​,21​), then α+β\alpha+\betaα+β is equal to
  1. (A)37
  2. (B)46
  3. (C)58
  4. (D)72

Correct answer: (C)

Step-by-step solution →
Q26·MathematicsSingle correctJEE Main 2025
Let the ellipse E1:x2a2+y2b2=1E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1E1​:a2x2​+b2y2​=1, a>ba>ba>b and E2:x2A2+y2B2=1E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1E2​:A2x2​+B2y2​=1, A<BA<BA<B have same eccentricity 13\dfrac{1}{\sqrt{3}}3​1​. Let the product of their lengths of latus rectums be 323\dfrac{32}{\sqrt{3}}3​32​, and the distance between the foci of E1E_1E1​ be 4. If E1E_1E1​ and E2E_2E2​ meet at A, B, C and D, then the area of the quadrilateral ABCD equals:
  1. (A)666\sqrt{6}66​
  2. (B)1865\dfrac{18\sqrt{6}}{5}5186​​
  3. (C)1265\dfrac{12\sqrt{6}}{5}5126​​
  4. (D)2465\dfrac{24\sqrt{6}}{5}5246​​

Correct answer: (D)

Step-by-step solution →
Q27·MathematicsSingle correctJEE Main 2025
If the midpoint of a chord of the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=19x2​+4y2​=1 is (2,43)\left(\sqrt{2},\frac{4}{3}\right)(2​,34​), and the length of the chord is 2α3\frac{2\sqrt{\alpha}}{3}32α​​, then α\alphaα is:
  1. (A)18
  2. (B)22
  3. (C)26
  4. (D)20

Correct answer: (B)

Step-by-step solution →
Q28·MathematicsIntegerJEE Main 2025
Let E1:x29+y24=1E_1:\frac{x^2}{9}+\frac{y^2}{4}=1E1​:9x2​+4y2​=1 be an ellipse. Ellipses EiE_iEi​'s are constructed such that their centres and eccentricities are same as that of E1E_1E1​, and the length of minor axis of EiE_iEi​ is the length of major axis of Ei+1E_{i+1}Ei+1​ (i≥1)(i\ge1)(i≥1). If AiA_iAi​ is the area of the ellipse EiE_iEi​, then 5π(∑i=1∞Ai)\frac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right)π5​(∑i=1∞​Ai​), is equal to ____.

Correct answer: 54

Step-by-step solution →
Q29·MathematicsSingle correctJEE Main 2025
Let the product of the focal distances of the point (3,12)\left(\sqrt3,\dfrac{1}{2}\right)(3​,21​) on the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1a2x2​+b2y2​=1, (a>b)(a>b)(a>b), be 74\dfrac{7}{4}47​. Then the absolute difference of the eccentricities of two such ellipses is
  1. (A)3−2232\dfrac{3-2\sqrt2}{3\sqrt2}32​3−22​​
  2. (B)1−32\dfrac{1-\sqrt3}{\sqrt2}2​1−3​​
  3. (C)3−2223\dfrac{3-2\sqrt2}{2\sqrt3}23​3−22​​
  4. (D)1−23\dfrac{1-\sqrt2}{\sqrt3}3​1−2​​

Correct answer: (C)

Step-by-step solution →
Q30·MathematicsSingle correctJEE Main 2025
The equation of the chord, of the ellipse x225+y216=1\dfrac{x^2}{25}+\dfrac{y^2}{16}=125x2​+16y2​=1, whose mid-point is (3,1)(3,1)(3,1) is :
  1. (A)48x+25y=16948x+25y=16948x+25y=169
  2. (B)4x+122y=1344x+122y=1344x+122y=134
  3. (C)25x+101y=17625x+101y=17625x+101y=176
  4. (D)5x+16y=315x+16y=315x+16y=31

Correct answer: (A)

Step-by-step solution →
Q31·MathematicsSingle correctJEE Main 2025
The length of the chord of the ellipse x24+y22=1\dfrac{x^2}{4}+\dfrac{y^2}{2}=14x2​+2y2​=1, whose mid-point is (1,12)\left(1,\dfrac{1}{2}\right)(1,21​), is :
  1. (A)2315\dfrac{2}{3}\sqrt{15}32​15​
  2. (B)5315\dfrac{5}{3}\sqrt{15}35​15​
  3. (C)1315\dfrac{1}{3}\sqrt{15}31​15​
  4. (D)15\sqrt{15}15​

Correct answer: (A)

Step-by-step solution →
Q32·MathematicsSingle correctJEE Advanced 2024
Consider the ellipse x29+y24=1\frac{x^{2}}{9} + \frac{y^{2}}{4} = 19x2​+4y2​=1. Let S(p,q)S(p, q)S(p,q) be a point in the first quadrant such that p29+q24>1\frac{p^{2}}{9} + \frac{q^{2}}{4} > 19p2​+4q2​>1. Two tangents are drawn from SSS to the ellipse, of which one meets the ellipse at one end point of the minor axis and the other meets the ellipse at a point TTT in the fourth quadrant. Let RRR be the vertex of the ellipse with positive x-coordinate and OOO be the centre of the ellipse. If the area of the triangle △ORT\triangle ORT△ORT is 32\frac{3}{2}23​, then which of the following options is correct ?
  1. (A)q=2,p=33q = 2, p = 3\sqrt{3}q=2,p=33​
  2. (B)q=2,p=43q = 2, p = 4\sqrt{3}q=2,p=43​
  3. (C)q=1,p=53q = 1, p = 5\sqrt{3}q=1,p=53​
  4. (D)q=1,p=63q = 1, p = 6\sqrt{3}q=1,p=63​

Correct answer: (A)

Step-by-step solution →
Q33·MathematicsSingle correctJEE Main 2024
Let the line 2x+3y−k=02x + 3y - k = 02x+3y−k=0, k>0k > 0k>0, intersect the x-axis and y-axis at the points A and B, respectively. If the equation of the circle having the line segment AB as a diameter is x2+y2−3x−2y=0x^2 + y^2 - 3x - 2y = 0x2+y2−3x−2y=0 and the length of the latus rectum of the ellipse x2+9y2=k2x^2 + 9y^2 = k^2x2+9y2=k2 is mn\dfrac{m}{n}nm​, where m and n are coprime, then 2m+n2m + n2m+n is equal to:
  1. (A)101010
  2. (B)111111
  3. (C)131313
  4. (D)121212

Correct answer: (B)

Step-by-step solution →
Q34·MathematicsSingle correctJEE Main 2024
Let P be a point on the ellipse x29+y24=1\dfrac{x^2}{9}+\dfrac{y^2}{4}=19x2​+4y2​=1. Let the line passing through P and parallel to y-axis meet the circle x2+y2=9x^2+y^2=9x2+y2=9 at point Q such that P and Q are on the same side of the x-axis. Then, the eccentricity of the locus of the point R on PQ such that PR:RQ=4:3PR:RQ=4:3PR:RQ=4:3 as P moves on the ellipse, is:
  1. (A)1119\dfrac{11}{19}1911​
  2. (B)1321\dfrac{13}{21}2113​
  3. (C)13923\dfrac{\sqrt{139}}{23}23139​​
  4. (D)137\dfrac{\sqrt{13}}{7}713​​

Correct answer: (D)

Step-by-step solution →
Q35·MathematicsSingle correctJEE Main 2024
Let x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1a2x2​+b2y2​=1, a>ba>ba>b, be an ellipse, whose eccentricity is 12\dfrac{1}{\sqrt2}2​1​ and the length of the latus rectum is 14\sqrt{14}14​. Then the square of the eccentricity of x2a2−y2b2=1\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1a2x2​−b2y2​=1 is:
  1. (A)333
  2. (B)72\dfrac7227​
  3. (C)32\dfrac3223​
  4. (D)52\dfrac5225​

Correct answer: (C)

Step-by-step solution →
Q36·MathematicsSingle correctJEE Main 2024
For 0<θ<π20<\theta<\dfrac\pi20<θ<2π​, if the eccentricity of the hyperbola x2−y2csc⁡2θ=5x^2-y^2\csc^2\theta=5x2−y2csc2θ=5 is 7\sqrt77​ times the eccentricity of the ellipse x2csc⁡2θ+y2=5x^2\csc^2\theta+y^2=5x2csc2θ+y2=5, then the value of θ\thetaθ is:
  1. (A)π6\dfrac\pi66π​
  2. (B)5π12\dfrac{5\pi}{12}125π​
  3. (C)π3\dfrac\pi33π​
  4. (D)π4\dfrac\pi44π​

Correct answer: (C)

Step-by-step solution →
Q37·MathematicsNumericalJEE Main 2024
Let the foci and length of the latus rectum of an ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1a2x2​+b2y2​=1, a>ba>ba>b be (±5,0)(\pm 5,0)(±5,0) and 50\sqrt{50}50​ respectively. Then, the square of the eccentricity of the hyperbola x2b2−y2a2b2=1\dfrac{x^2}{b^2}-\dfrac{y^2}{a^2 b^2}=1b2x2​−a2b2y2​=1 equals ______.

Correct answer: 51

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Q38·MathematicsSingle correctJEE Main 2024
If the foci of a hyperbola are same as that of the ellipse x29+y225=1\dfrac{x^2}{9}+\dfrac{y^2}{25}=19x2​+25y2​=1 and the eccentricity of the hyperbola is 158\dfrac{15}{8}815​ times the eccentricity of the ellipse, then the smaller focal distance of the point (2,14325)\left(\sqrt{2},\dfrac{14}{3}\sqrt{\dfrac{2}{5}}\right)(2​,314​52​​) on the hyperbola, is equal to
  1. (A)725−837\sqrt{\dfrac{2}{5}}-\dfrac{8}{3}752​​−38​
  2. (B)1425−4314\sqrt{\dfrac{2}{5}}-\dfrac{4}{3}1452​​−34​
  3. (C)1425−16314\sqrt{\dfrac{2}{5}}-\dfrac{16}{3}1452​​−316​
  4. (D)725+837\sqrt{\dfrac{2}{5}}+\dfrac{8}{3}752​​+38​

Correct answer: (A)

Step-by-step solution →
Q39·MathematicsSingle correctJEE Main 2024
Let A(α,0)A(\alpha,0)A(α,0) and B(0,β)B(0,\beta)B(0,β) be the points on the line 5x+7y=505x+7y=505x+7y=50. Let the point PPP divide the line segment ABABAB internally in the ratio 7:37:37:3. Let 3x−25=03x-25=03x−25=0 be a directrix of the ellipse E:x2a2+y2b2=1E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1E:a2x2​+b2y2​=1 and the corresponding focus be SSS. If from SSS, the perpendicular on the x-axis passes through PPP, then the length of the latus rectum of EEE is equal to:
  1. (A)253\dfrac{25}{3}325​
  2. (B)329\dfrac{32}{9}932​
  3. (C)259\dfrac{25}{9}925​
  4. (D)325\dfrac{32}{5}532​

Correct answer: (D)

Step-by-step solution →
Q40·MathematicsSingle correctJEE Main 2024
If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is:
  1. (A)53\dfrac{\sqrt5}{3}35​​
  2. (B)32\dfrac{\sqrt3}{2}23​​
  3. (C)13\dfrac{1}{\sqrt3}3​1​
  4. (D)25\dfrac{2}{\sqrt5}5​2​

Correct answer: (D)

Step-by-step solution →
Q41·MathematicsNumericalJEE Main 2024
If the points of intersection of two distinct conics x2+y2=4bx^2+y^2=4bx2+y2=4b and x216+y2b2=1\dfrac{x^2}{16}+\dfrac{y^2}{b^2}=116x2​+b2y2​=1 lie on the curve y2=3x2y^2=3x^2y2=3x2, then 333\sqrt{3}33​ times the area of the rectangle formed by the intersection points is ______.

Correct answer: 432

Step-by-step solution →
Q42·MathematicsSingle correctJEE Main 2024
Let e1e_1e1​ be the eccentricity of the hyperbola x216−y29=1\dfrac{x^2}{16}-\dfrac{y^2}{9}=116x2​−9y2​=1 and e2e_2e2​ be the eccentricity of the ellipse x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1a2x2​+b2y2​=1, a>ba>ba>b, which passes through the foci of the hyperbola. If e1e2=1e_1 e_2=1e1​e2​=1, then the length of the chord of the ellipse parallel to the x-axis and passing through (0,2)(0,2)(0,2) is :
  1. (A)454\sqrt545​
  2. (B)853\dfrac{8\sqrt5}{3}385​​
  3. (C)1053\dfrac{10\sqrt5}{3}3105​​
  4. (D)353\sqrt535​

Correct answer: (C)

Step-by-step solution →
Q43·MathematicsSingle correctJEE Main 2024
The length of the chord of the ellipse x225+y216=1\dfrac{x^2}{25}+\dfrac{y^2}{16}=125x2​+16y2​=1, whose mid point is (1,25)\left(1,\dfrac{2}{5}\right)(1,52​), is equal to:
  1. (A)16915\dfrac{\sqrt{1691}}{5}51691​​
  2. (B)20095\dfrac{\sqrt{2009}}{5}52009​​
  3. (C)17415\dfrac{\sqrt{1741}}{5}51741​​
  4. (D)15415\dfrac{\sqrt{1541}}{5}51541​​

Correct answer: (A)

Step-by-step solution →
Q44·MathematicsMultiple correctJEE Advanced 2023
Let T1T_1T1​ and T2T_2T2​ be two distinct common tangents to the ellipse E:x26+y23=1E : \frac{x^2}{6} + \frac{y^2}{3} = 1E:6x2​+3y2​=1 and the parabola P:y2=12xP : y^2 = 12xP:y2=12x. Suppose that the tangent T1T_1T1​ touches PPP and EEE at the points A1A_1A1​ and A2A_2A2​, respectively and the tangent T2T_2T2​ touches PPP and EEE at the points A4A_4A4​ and A3A_3A3​ , respectively. Then which of the following statements is(are) true?
  1. (A)The area of the quadrilateral A1A2A3A4A_1A_2A_3A_4A1​A2​A3​A4​ is 35 square units
  2. (B)The area of the quadrilateral A1A2A3A4A_1A_2A_3A_4A1​A2​A3​A4​ is 36 square units
  3. (C)The tangents T1T_1T1​ and T2T_2T2​ meet the x -axis at the point (−3,0)(-3, 0)(−3,0)
  4. (D)The tangents T1T_1T1​ and T2T_2T2​ meet the x -axis at the point (−6,0)(-6, 0)(−6,0)

Correct answer: (A), (C)

Step-by-step solution →
Q45·MathematicsNumericalJEE Main 2023
Let an ellipse with centre (1,0)(1, 0)(1,0) and latus rectum of length 12\frac{1}{2}21​ have its major axis along xxx-axis. If its minor axis subtends an angle 60∘60^\circ60∘ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to____

Correct answer: 9

Step-by-step solution →
Q46·MathematicsSingle correctJEE Main 2023
Let the tangent and normal at the point (33,1)(3\sqrt3,1)(33​,1) on the ellipse x236+y24=1\dfrac{x^2}{36}+\dfrac{y^2}{4}=136x2​+4y2​=1 meet the yyy-axis at the points AAA and BBB respectively. Let the circle CCC be drawn taking ABABAB as a diameter and the line x=25x=2\sqrt5x=25​ intersect CCC at the points PPP and QQQ. If the tangents at the points PPP and QQQ on the circle intersect at the point (α,0)(\alpha,0)(α,0) and (β,0)(\beta,0)(β,0), then α2−β2\alpha^2-\beta^2α2−β2 is equal to:
  1. (A)3145\dfrac{314}{5}5314​
  2. (B)3045\dfrac{304}{5}5304​
  3. (C)606060
  4. (D)616161

Correct answer: (B)

Step-by-step solution →
Q47·MathematicsSingle correctJEE Main 2023
Let P(237,67)\left(\dfrac{2\sqrt3}{\sqrt7},\dfrac{6}{\sqrt7}\right)(7​23​​,7​6​), Q, R and S be four points on the ellipse 9x2+4y2=369x^2+4y^2=369x2+4y2=36. Let PQ and RS be mutually perpendicular and pass through the origin. If 1(PQ)2+1(RS)2=pq\dfrac{1}{(PQ)^2}+\dfrac{1}{(RS)^2}=\dfrac{p}{q}(PQ)21​+(RS)21​=qp​, where p and q are coprime, then p+qp+qp+q is equal to
  1. (A)143
  2. (B)157
  3. (C)137
  4. (D)147

Correct answer: (B)

Step-by-step solution →
Q48·MathematicsSingle correctJEE Main 2023
If the radius of the largest circle with centre (2,0)(2,0)(2,0) inscribed in the ellipse x2+4y2=36x^2+4y^2=36x2+4y2=36 is rrr, then 12r212r^212r2 is equal to
  1. (A)727272
  2. (B)115115115
  3. (C)929292
  4. (D)696969

Correct answer: (C)

Step-by-step solution →
Q49·MathematicsSingle correctJEE Main 2023
Consider ellipses Ek:kx2+k2y2=1E_k : kx^2 + k^2 y^2 = 1Ek​:kx2+k2y2=1, k=1,2,…,20k = 1, 2, \ldots, 20k=1,2,…,20. Let CkC_kCk​ be the circle which touches the four chords joining the end points (one on minor axis and another on major axis) of the ellipse EkE_kEk​. If rkr_krk​ is the radius of the circle CkC_kCk​, then the value of ∑k=1201rk2\sum_{k=1}^{20} \frac{1}{r_k^2}∑k=120​rk2​1​ is
  1. (A)3080
  2. (B)3210
  3. (C)3320
  4. (D)2870

Correct answer: (A)

Step-by-step solution →
Q50·MathematicsSingle correctJEE Main 2023
Let a circle of radius 444 be concentric to the ellipse 15x2+19y2=28515x^2+19y^2=28515x2+19y2=285. Then the common tangents are inclined to the minor axis of the ellipse at the angle:
  1. (A)π4\frac{\pi}{4}4π​
  2. (B)π3\frac{\pi}{3}3π​
  3. (C)π12\frac{\pi}{12}12π​
  4. (D)π6\frac{\pi}{6}6π​

Correct answer: (B)

Step-by-step solution →
Q51·MathematicsSingle correctJEE Main 2023
Let the ellipse E: x2+9y2=9x^2+9y^2=9x2+9y2=9 intersect the positive x- and y-axes at the points A and B respectively. Let the major axis of E be a diameter of the circle C. Let the line passing through A and B meet the circle C at the point P. If the area of the triangle with vertices A, P and the origin O is mn\dfrac{m}{n}nm​, where m and n are coprime, then m−nm-nm−n is equal to
  1. (A)181818
  2. (B)161616
  3. (C)171717
  4. (D)151515

Correct answer: (C)

Step-by-step solution →
Q52·MathematicsNumericalJEE Main 2023
Let the eccentricity of an ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1a2x2​+b2y2​=1 be reciprocal to that of the hyperbola 2x2−2y2=12x^2-2y^2=12x2−2y2=1. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectum of the ellipse is

Correct answer: 2

Step-by-step solution →
Q53·MathematicsSingle correctJEE Main 2023
In a group of 100 persons 75 speak English and 40 speak Hindi. Each person speaks at least one of the two languages. If the number of persons, who speak only English is aaa and the number of persons who speak only Hindi is bbb, then the eccentricity of the ellipse 25(x2b2+y2a2)=125\left(\frac{x^2}{b^2}+\frac{y^2}{a^2}\right)=125(b2x2​+a2y2​)=1 is equal to
  1. (A)31512\frac{3\sqrt{15}}{12}12315​​
  2. (B)11712\frac{\sqrt{117}}{12}12117​​
  3. (C)11912\frac{\sqrt{119}}{12}12119​​
  4. (D)12912\frac{\sqrt{129}}{12}12129​​

Correct answer: (C)

Step-by-step solution →
Q54·MathematicsNumericalJEE Main 2023
The line x=8x=8x=8 is the directrix of the ellipse E:x2a2+y2b2=1E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1E:a2x2​+b2y2​=1 with the corresponding focus (2,0)(2,0)(2,0). If the tangent to E at the point P in the first quadrant passes through the point (0,43)(0,4\sqrt3)(0,43​) and intersects the x-axis at Q, then (3PQ)2(3PQ)^2(3PQ)2 equal to

Correct answer: 39

Step-by-step solution →
Q55·MathematicsSingle correctJEE Main 2023
If the maximum distance of normal to the ellipse x24+y2b2=1, b<2\frac{x^2}{4}+\frac{y^2}{b^2}=1,\ b<24x2​+b2y2​=1, b<2, from the origin is 111, then the eccentricity of the ellipse is :
  1. (A)12\frac{1}{2}21​
  2. (B)34\frac{\sqrt{3}}{4}43​​
  3. (C)32\frac{\sqrt{3}}{2}23​​
  4. (D)12\frac{1}{\sqrt{2}}2​1​

Correct answer: (C)

Step-by-step solution →
Q56·MathematicsNumericalJEE Main 2023
Let a tangent to the curve 9x2+16y2=1449x^2+16y^2=1449x2+16y2=144 intersect the coordinate axes at the points AAA and BBB. Then, the minimum length of the line segment ABABAB is _______ .

Correct answer: 7

Step-by-step solution →
Q57·MathematicsNumericalJEE Main 2023
Let CCC be the largest circle centred at (2,0)(2,0)(2,0) and inscribed in the ellipse x236+y216=1\dfrac{x^2}{36}+\dfrac{y^2}{16}=136x2​+16y2​=1. If (1,α)(1,\alpha)(1,α) lies on CCC, then 10α210\alpha^210α2 is equal to _______ .

Correct answer: 118

Step-by-step solution →
Q58·MathematicsSingle correctJEE Advanced 2022
Consider the ellipse x24+y23=1\frac{x^{2}}{4} + \frac{y^{2}}{3} = 14x2​+3y2​=1. Let H(α,0)H\left(\alpha, 0\right)H(α,0), 0<α<20 < \alpha < 20<α<2, be a point. A straight line drawn through HHH parallel to yyy-axis crosses the ellipse and its auxiliary circle at points EEE and FFF respectively, in the first quadrant. The tangents to the ellipse at the point EEE intersects the positive xxx-axis at a point GGG. Suppose the straight line joining FFF and the origin makes an angle ϕ\phiϕ with the positive xxx-axis. The correct option is:
List-IList-II
I.If ϕ=π4\phi = \frac{\pi}{4}ϕ=4π​, then the area of the triangle FGHFGHFGH isP.(3−1)48\frac{\left(\sqrt{3} - 1\right)^{4}}{8}8(3​−1)4​
II.If ϕ=π3\phi = \frac{\pi}{3}ϕ=3π​, then the area of the triangle FGHFGHFGH isQ.111
III.If ϕ=π6\phi = \frac{\pi}{6}ϕ=6π​, then the area of the triangle FGHFGHFGH isR.34\frac{3}{4}43​
IV.If ϕ=π12\phi = \frac{\pi}{12}ϕ=12π​, then the area of the triangle FGHFGHFGH isS.123\frac{1}{2\sqrt{3}}23​1​
T.332\frac{3\sqrt{3}}{2}233​​
  1. (A)(I) → (R); (II) → (S); (III) → (Q); (IV) → (P)
  2. (B)(I) → (R); (II) → (T); (III) → (S); (IV) → (P)
  3. (C)(I) → (Q); (II) → (T); (III) → (S); (IV) → (P)
  4. (D)(I) → (Q); (II) → (S); (III) → (Q); (IV) → (P)

Correct answer: (C)

Step-by-step solution →
Q59·MathematicsNumericalJEE Main 2022
Let S={(x,y)∈N×N:9(x−3)2+16(y−4)2≤144}S = \left\{ (x, y) \in \mathbb{N} \times \mathbb{N} : 9(x - 3)^2 + 16(y - 4)^2 \le 144 \right\}S={(x,y)∈N×N:9(x−3)2+16(y−4)2≤144} and T={(x,y)∈R×R:(x−7)2+(y−4)2≤36}T = \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : (x - 7)^2 + (y - 4)^2 \le 36 \right\}T={(x,y)∈R×R:(x−7)2+(y−4)2≤36}. The n(S∩T)n(S \cap T)n(S∩T) is equal to______.

Correct answer: 27

Step-by-step solution →
Q60·MathematicsSingle correctJEE Main 2022
Let a line L pass through the point of intersection of the lines bx+10y−8=0bx + 10y - 8 = 0bx+10y−8=0 and 2x−3y=02x - 3y = 02x−3y=0, b∈R−{43}b \in R - \left\{\frac{4}{3}\right\}b∈R−{34​}. If the line L also passes through the point (1,1)(1, 1)(1,1) and touches the circle 17 (x2+y2)=1617\,(x^{2} + y^{2}) = 1617(x2+y2)=16, then the eccentricity of the ellipse x25+y2b2=1\frac{x^{2}}{5} + \frac{y^{2}}{b^{2}} = 15x2​+b2y2​=1 is :
  1. (A)25\frac{2}{\sqrt{5}}5​2​
  2. (B)35\sqrt{\frac{3}{5}}53​​
  3. (C)15\frac{1}{\sqrt{5}}5​1​
  4. (D)25\sqrt{\frac{2}{5}}52​​

Correct answer: (B)

Step-by-step solution →
Q61·MathematicsNumericalJEE Main 2022
Let the tangents at the points P and Q on the ellipse x22+y24=1\frac{x^{2}}{2}+\frac{y^{2}}{4}=12x2​+4y2​=1 meet at the point R(2, 22−2)R\left(\sqrt{2},\,2\sqrt{2}-2\right)R(2​,22​−2). If S is the focus of the ellipse on its negative major axis, then SP2+SQ2SP^{2} + SQ^{2}SP2+SQ2 is equal to

Correct answer: 13

Step-by-step solution →
Q62·MathematicsNumericalJEE Main 2022
For the hyperbola H : x2−y2=1x^{2} - y^{2} = 1x2−y2=1 and the ellipse E : x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1a2x2​+b2y2​=1, a > b > 0, let the (1) eccentricity of E be reciprocal of the eccentricity of H, and (2) the line y=52x+Ky = \sqrt{\frac{5}{2}}x + Ky=25​​x+K be a common tangent of E and H. Then 4(a2+b2)4(a^{2} + b^{2})4(a2+b2) is equal to ________.

Correct answer: 3

Step-by-step solution →
Q63·MathematicsNumericalJEE Main 2022
An ellipse E:x2a2+y2b2=1E:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1E:a2x2​+b2y2​=1 passes through the vertices of the hyperbola H:x249−y264=−1H:\frac{x^{2}}{49}-\frac{y^{2}}{64}=-1H:49x2​−64y2​=−1. Let the major and minor axes of the ellipse E coincide with the transverse and conjugate axes of the hyperbola H. Let the product of the eccentricities of E and H be 12\frac{1}{2}21​. If lll is the length of the latus rectum of the ellipse E, then the value of 113l113l113l is equal to ______.

Correct answer: 1552

Step-by-step solution →
Q64·MathematicsNumericalJEE Main 2022
If the length of the latus rectum of the ellipse x2+4y2+2x+8y−λ=0x^{2}+4y^{2}+2x+8y-\lambda=0x2+4y2+2x+8y−λ=0 is 4, and lll is the length of its major axis, then λ+l\lambda+lλ+l is equal to ______.

Correct answer: 75

Step-by-step solution →
Q65·MathematicsSingle correctJEE Main 2022
The acute angle between the pair of tangents drawn to the ellipse 2x2+3y2=52x^{2}+3y^{2}=52x2+3y2=5 from the point (1,3)(1,3)(1,3) is
  1. (A)tan⁡−1(1675)\tan^{-1}\left(\frac{16}{7\sqrt{5}}\right)tan−1(75​16​)
  2. (B)tan⁡−1(2475)\tan^{-1}\left(\frac{24}{7\sqrt{5}}\right)tan−1(75​24​)
  3. (C)tan⁡−1(3275)\tan^{-1}\left(\frac{32}{7\sqrt{5}}\right)tan−1(75​32​)
  4. (D)tan⁡−1(3+8535)\tan^{-1}\left(\frac{3+8\sqrt{5}}{35}\right)tan−1(353+85​​)

Correct answer: (B)

Step-by-step solution →
Q66·MathematicsSingle correctJEE Main 2022
Let PQ be a focal chord of the parabola y2=4xy^2 = 4xy2=4x such that it subtends an angle of π2\frac{\pi}{2}2π​ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E: x2a2+y2b2=1,a2>b2\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a^2 > b^2a2x2​+b2y2​=1,a2>b2. If e is the eccentricity of the ellipse E, then the value of 1e2\frac{1}{e^2}e21​ is equal to :
  1. (A)1+21 + \sqrt{2}1+2​
  2. (B)3+223 + 2\sqrt{2}3+22​
  3. (C)1+231 + 2\sqrt{3}1+23​
  4. (D)4+534 + 5\sqrt{3}4+53​

Correct answer: (B)

Step-by-step solution →
Q67·MathematicsSingle correctJEE Main 2022
Let the eccentricity of an ellipse x2a2+y2b2=1,a>b,\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b,a2x2​+b2y2​=1,a>b, be 14\frac{1}{4}41​. If this ellipse passes through the point (−425,3)\left(-4\sqrt{\frac{2}{5}}, 3\right)(−452​​,3), then a2+b2a^{2} + b^{2}a2+b2 is equal to :
  1. (A)29
  2. (B)31
  3. (C)32
  4. (D)34

Correct answer: (B)

Step-by-step solution →
Q68·MathematicsSingle correctJEE Main 2022
The locus of the mid point of the line segment joining the point (4, 3) and the points on the ellipse x2+2y2=4x^2+2y^2=4x2+2y2=4 is an ellipse with eccentricity :
  1. (A)32\frac{\sqrt{3}}{2}23​​
  2. (B)122\frac{1}{2\sqrt{2}}22​1​
  3. (C)12\frac{1}{\sqrt{2}}2​1​
  4. (D)12\frac{1}{2}21​

Correct answer: (C)

Step-by-step solution →
Q69·MathematicsNumericalJEE Main 2022
Let the common tangents to the curves 4(x2+y2)=94(x^{2} + y^{2}) = 94(x2+y2)=9 and y2=4xy^{2} = 4xy2=4x intersect at the point Q. Let an ellipse, centered at the origin O, has lengths of semi-minor and semi-major axes equal to OQ and 6, respectively. If eee and lll respectively denote the eccentricity and the length of the latus rectum of this ellipse, then le2\frac{l}{e^{2}}e2l​ is equal to____________.

Correct answer: 4

Step-by-step solution →
Q70·MathematicsSingle correctJEE Main 2022
If m is the slope of a common tangent to the curves x216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=116x2​+9y2​=1 and x2+y2=12x^2+y^2=12x2+y2=12, then 12m212m^212m2 is equal to :
  1. (A)6
  2. (B)9
  3. (C)10
  4. (D)12

Correct answer: (B)

Step-by-step solution →
Q71·MathematicsSingle correctJEE Main 2022
The line y=x+1y = x + 1y=x+1 meets the ellipse x24+y22=1\frac{x^{2}}{4} + \frac{y^{2}}{2} = 14x2​+2y2​=1 at two points P and Q. If r is the radius of the circle with PQ as diameter then (3r)2(3r)^{2}(3r)2 is equal to
  1. (A)20
  2. (B)12
  3. (C)11
  4. (D)8

Correct answer: (A)

Step-by-step solution →
Q72·MathematicsSingle correctJEE Main 2022
Let the maximum area of the triangle that can be inscribed in the ellipse x2a2+y24=1\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{4}=1a2x2​+4y2​=1, a>2a > 2a>2, having one of its vertices at one end of the major axis of the ellipse and one of its sides parallel to the y-axis, be 636\sqrt{3}63​. Then the eccentricity of the ellispe is :
  1. (A)32\dfrac{\sqrt{3}}{2}23​​
  2. (B)12\dfrac{1}{2}21​
  3. (C)12\dfrac{1}{\sqrt{2}}2​1​
  4. (D)34\dfrac{\sqrt{3}}{4}43​​

Correct answer: (A)

Step-by-step solution →
Q73·MathematicsIntegerJEE Advanced 2021
Let E be the ellipse x216+y29=1\frac{x^2}{16} + \frac{y^2}{9} = 116x2​+9y2​=1. For any three distinct points P, Q and Q' on E, let M (P, Q) be the mid-point of the line segment joining P and Q, and M (P, Q') be the mid-point of the line segment joining P and Q'. Then the maximum possible value of the distance between M(P, Q) and M(P, Q'), as P, Q and Q' vary on E, is _____.

Correct answer: 4

Step-by-step solution →
Q74·MathematicsSingle correctJEE Main 2021
Let θ be the acute angle between the tangents to the ellipse x29+y21=1\frac{x^{2}}{9} + \frac{y^{2}}{1} = 19x2​+1y2​=1 and the circle x2+y2=3x^{2} + y^{2} = 3x2+y2=3 at their point of intersection in the first quadrant. Then tanθ is equal to :
  1. (A)523\frac{5}{2\sqrt{3}}23​5​
  2. (B)23\frac{2}{\sqrt{3}}3​2​
  3. (C)43\frac{4}{\sqrt{3}}3​4​
  4. (D)2

Correct answer: (B)

Step-by-step solution →
Q75·MathematicsSingle correctJEE Main 2021
The locus of mid-points of the line segments joining (−3,−5)(-3, -5)(−3,−5) and the points on the ellipse x24+y29=1\frac{x^2}{4} + \frac{y^2}{9} = 14x2​+9y2​=1 is :
  1. (A)9x2+4y2+18x+8y+145=09x^2 + 4y^2 + 18x + 8y + 145 = 09x2+4y2+18x+8y+145=0
  2. (B)36x2+16y2+90x+56y+145=036x^2 + 16y^2 + 90x + 56y + 145 = 036x2+16y2+90x+56y+145=0
  3. (C)36x2+16y2+108x+80y+145=036x^2 + 16y^2 + 108x + 80y + 145 = 036x2+16y2+108x+80y+145=0
  4. (D)36x2+16y2+72x+32y+145=036x^2 + 16y^2 + 72x + 32y + 145 = 036x2+16y2+72x+32y+145=0

Correct answer: (C)

Step-by-step solution →
Q76·MathematicsSingle correctJEE Main 2021
The line 12xcos⁡θ+5ysin⁡θ=6012x \cos\theta + 5y \sin\theta = 6012xcosθ+5ysinθ=60 is tangent to which of the following curves?
  1. (A)x2+y2=169x^{2} + y^{2} = 169x2+y2=169
  2. (B)144x2+25y2=3600144x^{2} + 25y^{2} = 3600144x2+25y2=3600
  3. (C)25x2+12y2=360025x^{2} + 12y^{2} = 360025x2+12y2=3600
  4. (D)x2+y2=60x^{2} + y^{2} = 60x2+y2=60

Correct answer: (B)

Step-by-step solution →
Q77·MathematicsSingle correctJEE Main 2021
On the ellipse x28+y24=1\frac{x^2}{8} + \frac{y^2}{4} = 18x2​+4y2​=1 let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x+2y=0x + 2y = 0x+2y=0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of (5−e2)(5 - e^2)(5−e2). A is :
  1. (A)6
  2. (B)12
  3. (C)14
  4. (D)24

Correct answer: (A)

Step-by-step solution →
Q78·MathematicsNumericalJEE Main 2021
Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its center at (3,−4), one focus at (4,−4) and one vertex at (5,−4) . If mx − y = 4, m > 0 is a tangent to the ellipse E, then the value of 5m2^22 is equal to……………

Correct answer: 3

Step-by-step solution →
Q79·MathematicsSingle correctJEE Main 2021
A ray of light through (2, 1) is reflected at a point P on the y-axis and then passes through the point (5, 3). If this reflected ray is the directrix of any ellipse with eccentricity 13\frac{1}{3}31​ and the distance of the nearer focus from this directrix is 853\frac{8}{\sqrt{53}}53​8​, then the equation of the other directrix can be :
  1. (A)2x−7y−39=02x - 7y - 39 = 02x−7y−39=0 or  2x−7y−7=0\ 2x - 7y - 7 = 0 2x−7y−7=0
  2. (B)11x+7y+8=011x + 7y + 8 = 011x+7y+8=0 or  11x+7y−15=0\ 11x + 7y - 15 = 0 11x+7y−15=0
  3. (C)2x−7y+29=02x - 7y + 29 = 02x−7y+29=0 or  2x−7y−7=0\ 2x - 7y - 7 = 0 2x−7y−7=0
  4. (D)11x−7y−8=011x - 7y - 8 = 011x−7y−8=0 or  11x+7y+15=0\ 11x + 7y + 15 = 0 11x+7y+15=0

Correct answer: (C)

Step-by-step solution →
Q80·MathematicsSingle correctJEE Main 2021
If a tangent to the ellipse x2+4y2=4x^{2}+4y^{2}=4x2+4y2=4 meets the tangents at the extremities of its major axis at B and C, then the circle with BC as diameter passes through the point :
  1. (A)(1,1)(1,1)(1,1)
  2. (B)(2,0)(\sqrt{2},0)(2​,0)
  3. (C)(3,0)(\sqrt{3},0)(3​,0)
  4. (D)(−1,1)(-1,1)(−1,1)

Correct answer: (C)

Step-by-step solution →
Q81·MathematicsSingle correctJEE Main 2021
Let an ellipse E:x2a2+y2b2=1,a2>b2E:\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1, a^{2}>b^{2}E:a2x2​+b2y2​=1,a2>b2, passes through (32,1)\left(\sqrt{\frac{3}{2}},1\right)(23​​,1) and has eccentricity 13\frac{1}{\sqrt{3}}3​1​. If a circle, centered at focus F(α,0),α>0F(\alpha,0), \alpha>0F(α,0),α>0, of E and radius 23\frac{2}{\sqrt{3}}3​2​, intersects E at two points P and Q, then PQ2PQ^{2}PQ2 is equal to :
  1. (A)163\frac{16}{3}316​
  2. (B)3
  3. (C)43\frac{4}{3}34​
  4. (D)83\frac{8}{3}38​

Correct answer: (A)

Step-by-step solution →
Q82·MathematicsSingle correctJEE Main 2021
Let E1:x2a2+y2b2=1E_1 : \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1E1​:a2x2​+b2y2​=1, a>ba > ba>b. Let E2E_2E2​ be another ellipse such that it touches the end points of major axis of E1E_1E1​ and the foci of E2E_2E2​ are the end points of minor axis of E1E_1E1​. If E1E_1E1​ and E2E_2E2​ have same eccentricities, then its value is :
  1. (A)−1+52\dfrac{-1 + \sqrt{5}}{2}2−1+5​​
  2. (B)−1+32\dfrac{-1 + \sqrt{3}}{2}2−1+3​​
  3. (C)−1+82\dfrac{-1 + \sqrt{8}}{2}2−1+8​​
  4. (D)−1+62\dfrac{-1 + \sqrt{6}}{2}2−1+6​​

Correct answer: (A)

Step-by-step solution →
Q83·MathematicsSingle correctJEE Main 2021
Let a tangent be drawn to the ellipse x227+y2=1\frac{x^{2}}{27} + y^{2} = 127x2​+y2=1 at (33cos⁡θ,sin⁡θ)\left(3\sqrt{3}\cos\theta, \sin\theta\right)(33​cosθ,sinθ) where θ∈(0,π2)\theta \in \left(0, \frac{\pi}{2}\right)θ∈(0,2π​). Then the value of θ such that the sum of intercepts on axes made by this tangent is minimum is equal to :
  1. (A)π8\frac{\pi}{8}8π​
  2. (B)π4\frac{\pi}{4}4π​
  3. (C)π6\frac{\pi}{6}6π​
  4. (D)π3\frac{\pi}{3}3π​

Correct answer: (C)

Step-by-step solution →
Q84·MathematicsSingle correctJEE Main 2021
Let L be a tangent line to the parabola y2=4x−20y^{2} = 4x - 20y2=4x−20 at (6,2)(6, 2)(6,2). If L is also a tangent to the ellipse x22+y2b=1\frac{x^{2}}{2}+\frac{y^{2}}{b} = 12x2​+by2​=1, then the value of bbb is equal to :
  1. (A)111111
  2. (B)141414
  3. (C)161616
  4. (D)202020

Correct answer: (B)

Step-by-step solution →
Q85·MathematicsSingle correctJEE Main 2021
If the point of intersections of the ellipse x216+y2b2=1\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=116x2​+b2y2​=1 and the circle x2+y2=4bx^{2}+y^{2}=4bx2+y2=4b, b>4b > 4b>4 lie on the curve y2=3x2y^{2}=3x^{2}y2=3x2, then b is equal to:
  1. (A)121212
  2. (B)555
  3. (C)666
  4. (D)101010

Correct answer: (A)

Step-by-step solution →
Q86·MathematicsNumericalJEE Main 2021
Let L be a common tangent line to the curves 4x2+9y2=364x^{2} + 9y^{2} = 364x2+9y2=36 and (2x)2+(2y)2=31(2x)^{2} + (2y)^{2} = 31(2x)2+(2y)2=31. Then the square of the slope of the line L is ________________.

Correct answer: 3

Step-by-step solution →
Q87·MathematicsSingle correctJEE Main 2021
For which of the following curves, the line x+3y=23x+\sqrt{3}y=2\sqrt{3}x+3​y=23​ is the tangent at the point (332,12)\left(\frac{3\sqrt{3}}{2},\frac{1}{2}\right)(233​​,21​)?
  1. (A)x2+9y2=9x^{2}+9y^{2}=9x2+9y2=9
  2. (B)2x2−18y2=92x^{2}-18y^{2}=92x2−18y2=9
  3. (C)y2=163xy^{2}=\frac{1}{6\sqrt{3}}xy2=63​1​x
  4. (D)x2+y2=7x^{2}+y^{2}=7x2+y2=7

Correct answer: (A)

Step-by-step solution →
Q88·MathematicsSingle correctJEE Advanced 2020
Let a, b and λ\lambdaλ be positive real numbers. Suppose P is an end point of the latus rectum of the parabola y2=4λxy^{2} = 4\lambda xy2=4λx, and suppose the ellipse x2a2+y2b2=1\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1a2x2​+b2y2​=1 passes through the point P. If the tangents to the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is
  1. (A)12\frac{1}{\sqrt{2}}2​1​
  2. (B)12\frac{1}{2}21​
  3. (C)13\frac{1}{3}31​
  4. (D)25\frac{2}{5}52​

Correct answer: (A)

Step-by-step solution →
Q89·MathematicsSingle correctJEE Main 2020
If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies:
  1. (A)e4+2e2−1=0e^4 + 2e^2 - 1 = 0e4+2e2−1=0
  2. (B)e2+e−1=0e^2 + e - 1 = 0e2+e−1=0
  3. (C)e2+2e−1=0e^2 + 2e - 1 = 0e2+2e−1=0
  4. (D)e4+e2−1=0e^4 + e^2 - 1 = 0e4+e2−1=0

Correct answer: (D)

Step-by-step solution →
Q90·MathematicsSingle correctJEE Main 2020
Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse, x24+y22=1\dfrac{x^2}{4} + \dfrac{y^2}{2} = 14x2​+2y2​=1 from any of its foci?
  1. (A)(−2,3)(-2, \sqrt{3})(−2,3​)
  2. (B)(−1,2)(-1, \sqrt{2})(−1,2​)
  3. (C)(−1,3)(-1, \sqrt{3})(−1,3​)
  4. (D)(1,2)(1, 2)(1,2)

Correct answer: (C)

Step-by-step solution →
Q91·MathematicsSingle correctJEE Main 2020
If the co-ordinates of two points A and B are (7,0)(\sqrt{7},0)(7​,0) and (−7,0)(-\sqrt{7},0)(−7​,0) respectively and P is any point on the conic 9x2+16y2=1449x^{2}+16y^{2}=1449x2+16y2=144, then PA + PB is equal to:
  1. (A)8
  2. (B)16
  3. (C)9
  4. (D)6

Correct answer: (A)

Step-by-step solution →
Q92·MathematicsSingle correctJEE Main 2020
Let x2a2+y2b2=1 (a>b)\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1\,(a > b)a2x2​+b2y2​=1(a>b) be given ellipse, length of whose latus rectum is 10. If its eccentricity is the maximum value of the function, ϕ(t)=512+t−t2\phi(t) = \frac{5}{12} + t - t^{2}ϕ(t)=125​+t−t2, then a2+b2a^{2} + b^{2}a2+b2 is equal to:
  1. (A)126126126
  2. (B)135135135
  3. (C)116116116
  4. (D)145145145

Correct answer: (A)

Step-by-step solution →
Q93·MathematicsSingle correctJEE Main 2020
Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 12\frac{1}{2}21​. If P(1, β), β > 0 is a point on this ellipse, then the equation of the normal to it at P is:
  1. (A)4x − 2y = 1
  2. (B)4x − 3y = 2
  3. (C)7x − 4y = 1
  4. (D)8x − 2y = 5

Correct answer: (A)

Step-by-step solution →
Q94·MathematicsSingle correctJEE Main 2020
Let e1e_1e1​ and e2e_2e2​ be the eccentricities of the ellipse, x225+y2b2=1(b<5)\frac{x^2}{25} + \frac{y^2}{b^2} = 1 \left( b < 5 \right)25x2​+b2y2​=1(b<5) and the hyperbola, x216−y2b2=1\frac{x^2}{16} - \frac{y^2}{b^2} = 116x2​−b2y2​=1 respectively satisfying e1e2=1e_1 e_2 = 1e1​e2​=1. If α\alphaα and β\betaβ are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair (α,β)\left( \alpha, \beta \right)(α,β) is equal to :
  1. (A)(8,10)(8, 10)(8,10)
  2. (B)(245,10)\left( \frac{24}{5}, 10 \right)(524​,10)
  3. (C)(203,12)\left( \frac{20}{3}, 12 \right)(320​,12)
  4. (D)(8,12)(8, 12)(8,12)

Correct answer: (A)

Step-by-step solution →
Q95·MathematicsSingle correctJEE Main 2020
The length of the minor axis (along y − axis) of an ellipse in the standard form is 43\frac{4}{\sqrt{3}}3​4​. If this ellipse touches the line, x+6y=8x + 6y = 8x+6y=8; then its eccentricity is:
  1. (A)13113\frac{1}{3}\sqrt{\frac{11}{3}}31​311​​
  2. (B)1253\frac{1}{2}\sqrt{\frac{5}{3}}21​35​​
  3. (C)12113\frac{1}{2}\sqrt{\frac{11}{3}}21​311​​
  4. (D)56\sqrt{\frac{5}{6}}65​​

Correct answer: (C)

Step-by-step solution →
Q96·MathematicsSingle correctJEE Main 2020
If e1e_1e1​ and e2e_2e2​ are the eccentricities of the ellipse, x218+y24=1\dfrac{x^2}{18}+\dfrac{y^2}{4}=118x2​+4y2​=1 and the hyperbola, x29−y24=1\dfrac{x^2}{9}-\dfrac{y^2}{4}=19x2​−4y2​=1 respectively and (e1,e2)(e_1,e_2)(e1​,e2​) is a point on the ellipse, 15x2+3y2=k15x^2+3y^2=k15x2+3y2=k, then k is equal to:
  1. (A)16
  2. (B)14
  3. (C)17
  4. (D)15

Correct answer: (A)

Step-by-step solution →
Q97·MathematicsSingle correctJEE Main 2020
Let the line y=mxy=mxy=mx and the ellipse 2x2+y2=12x^{2}+y^{2}=12x2+y2=1 intersect at a point P in the first quadrant. If the normal to this ellipse at P meets the co-ordinate axes at (−132,0)\left(-\dfrac{1}{3\sqrt{2}},0\right)(−32​1​,0) and (0,β)(0,\beta)(0,β) then β\betaβ is equal to:
  1. (A)223\dfrac{2\sqrt{2}}{3}322​​
  2. (B)23\dfrac{2}{3}32​
  3. (C)23\dfrac{2}{\sqrt{3}}3​2​
  4. (D)23\dfrac{\sqrt{2}}{3}32​​

Correct answer: (D)

Step-by-step solution →
Q98·MathematicsSingle correctJEE Main 2020
If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is:
  1. (A)3\sqrt{3}3​
  2. (B)323\sqrt{2}32​
  3. (C)32\frac{3}{\sqrt{2}}2​3​
  4. (D)232\sqrt{3}23​

Correct answer: (B)

Step-by-step solution →
Q99·MathematicsSingle correctJEE Main 2020
If 3x+4y=1223x+4y=12\sqrt{2}3x+4y=122​ is a tangent to the ellipse x2a2+y29=1\frac{x^2}{a^2}+\frac{y^2}{9}=1a2x2​+9y2​=1 for some a∈Ra \in Ra∈R, then the distance between the foci of the ellipse is:
  1. (A)222\sqrt{2}22​
  2. (B)272\sqrt{7}27​
  3. (C)444
  4. (D)252\sqrt{5}25​

Correct answer: (B)

Step-by-step solution →
Q100·MathematicsMultiple correctJEE Advanced 2019
Define the collections {E1,E2,E3,.......}\{E_1, E_2, E_3, .......\}{E1​,E2​,E3​,.......} of ellipses and {R1,R2,R3,...}\{R_1, R_2, R_3, ...\}{R1​,R2​,R3​,...} of rectangles as follows: E1:x29+y24=1E_1 : \frac{x^2}{9} + \frac{y^2}{4} = 1E1​:9x2​+4y2​=1; R1R_1R1​ : rectangle of largest area, with sides parallel to the axes, inscribed in E1E_1E1​; EnE_nEn​ : ellipse x2an2+y2bn2=1\frac{x^2}{a_n^2} + \frac{y^2}{b_n^2} = 1an2​x2​+bn2​y2​=1 of largest area inscribed in Rn−1R_{n-1}Rn−1​, n>1n > 1n>1; RnR_nRn​ : rectangle of largest area, with sides parallel to the axes, inscribed in EnE_nEn​, n>1n > 1n>1. Then which of the following options is/are correct?
  1. (A)∑n=1N(area of Rn)<24\sum_{n=1}^{N} \left(\text{area of } R_n\right) < 24∑n=1N​(area of Rn​)<24, for each positive integer N
  2. (B)The distance of a focus from the centre in E9E_9E9​ is 532\frac{\sqrt{5}}{32}325​​
  3. (C)The eccentricities of E18E_{18}E18​ and E19E_{19}E19​ are NOT equal
  4. (D)The length of latus rectum of E9E_9E9​ is 16\frac{1}{6}61​

Correct answer: (A), (D)

Step-by-step solution →
Q101·MathematicsSingle correctJEE Main 2019
If the normal to the ellipse 3x2+4y2=123x^{2} + 4y^{2} = 123x2+4y2=12 at a point P on it is parallel to the line, 2x+y=42x + y = 42x+y=4 and the tangent to the ellipse at P passes through Q(4, 4) then PQ is equal to :
  1. (A)1572\dfrac{\sqrt{157}}{2}2157​​
  2. (B)552\dfrac{5\sqrt{5}}{2}255​​
  3. (C)2212\dfrac{\sqrt{221}}{2}2221​​
  4. (D)612\dfrac{\sqrt{61}}{2}261​​

Correct answer: (B)

Step-by-step solution →
Q102·MathematicsSingle correctJEE Main 2019
An ellipse, with foci at (0, 2) and (0, −2-2−2) and minor axis of length 4, passes through which of the following points?
  1. (A)(2,2)\left(2, \sqrt{2}\right)(2,2​)
  2. (B)(2,22)\left(2, 2\sqrt{2}\right)(2,22​)
  3. (C)(1,22)\left(1, 2\sqrt{2}\right)(1,22​)
  4. (D)(2,2)\left(\sqrt{2}, 2\right)(2​,2)

Correct answer: (D)

Step-by-step solution →
Q103·MathematicsSingle correctJEE Main 2019
The tangent and normal to the ellipse 3x2+5y2=323x^{2} + 5y^{2} = 323x2+5y2=32 at the point P(2,2)P(2, 2)P(2,2) meet the x-axis at Q and R, respectively. Then the area(in sq. units) of the triangle PQR is
  1. (A)3415\dfrac{34}{15}1534​
  2. (B)6815\dfrac{68}{15}1568​
  3. (C)143\dfrac{14}{3}314​
  4. (D)163\dfrac{16}{3}316​

Correct answer: (B)

Step-by-step solution →
Q104·MathematicsSingle correctJEE Main 2019
If the tangents on the ellipse 4x2+y2=84x^{2}+y^{2}=84x2+y2=8 at the points (1, 2) and (a, b) are perpendicular to each other, then a2a^{2}a2 is equal to:
  1. (A)217\dfrac{2}{17}172​
  2. (B)417\dfrac{4}{17}174​
  3. (C)6417\dfrac{64}{17}1764​
  4. (D)12817\dfrac{128}{17}17128​

Correct answer: (A)

Step-by-step solution →
Q105·MathematicsSingle correctJEE Main 2019
Let O (0, 0) and A (0, 1) be two fixed points. Then the locus of a point P such that the perimeter of △AOP\triangle AOP△AOP, is 4, is:
  1. (A)9x2−8y2+8y=169x^{2}-8y^{2}+8y=169x2−8y2+8y=16
  2. (B)8x2+9y2−9y=188x^{2}+9y^{2}-9y=188x2+9y2−9y=18
  3. (C)9x2+8y2−8y=169x^{2}+8y^{2}-8y=169x2+8y2−8y=16
  4. (D)8x2−9y2+9y=188x^{2}-9y^{2}+9y=188x2−9y2+9y=18

Correct answer: (C)

Step-by-step solution →
Q106·MathematicsSingle correctJEE Main 2019
In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at (0,53)(0, 5\sqrt{3})(0,53​), then the length of its latus rectum is:
  1. (A)6
  2. (B)5
  3. (C)8
  4. (D)10

Correct answer: (B)

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Q107·MathematicsSingle correctJEE Main 2019
Let S and S' be the foci of an ellipse and B be any one of the extremities of its minor axis. If △S′BS\triangle S'BS△S′BS is a right angled triangle with right angle at B and area (△S′BS)=8(\triangle S'BS) = 8(△S′BS)=8 sq. units, then the length of a latus rectum of the ellipse is :
  1. (A)4
  2. (B)222\sqrt{2}22​
  3. (C)424\sqrt{2}42​
  4. (D)2

Correct answer: (A)

Step-by-step solution →
Q108·MathematicsSingle correctJEE Main 2019
If tangents are drawn to the ellipse x2+2y2=2x^{2}+2y^{2}=2x2+2y2=2 at all points on the ellipse other than its four vertices than the mid points of the tangents intercepted between the coordinate axes lie on the curve:
  1. (A)14x2+12y2=1\frac{1}{4x^{2}}+\frac{1}{2y^{2}}=14x21​+2y21​=1
  2. (B)x24+y22=1\frac{x^{2}}{4}+\frac{y^{2}}{2}=14x2​+2y2​=1
  3. (C)12x2+14y2=1\frac{1}{2x^{2}}+\frac{1}{4y^{2}}=12x21​+4y21​=1
  4. (D)x22+y24=1\frac{x^{2}}{2}+\frac{y^{2}}{4}=12x2​+4y2​=1

Correct answer: (C)

Step-by-step solution →
Q109·MathematicsMultiple correctJEE Advanced 2018
Consider two straight lines, each of which is tangent to both the circle x2+y2=12x^{2} + y^{2} = \frac{1}{2}x2+y2=21​ and the parabola y2=4xy^{2} = 4xy2=4x. Let these lines intersect at the point Q. Consider the ellipse whose center is at the origin O(0, 0) and whose semi-major axis is OQ. If the length of the minor axis of this ellipse is 2\sqrt{2}2​, then which of the following statement(s) is (are) TRUE ?
  1. (A)For the ellipse, the eccentricity is 12\frac{1}{\sqrt{2}}2​1​ and the length of the latus rectum is 1
  2. (B)For the ellipse, the eccentricity is 12\frac{1}{2}21​ and the length of the latus rectum is 12\frac{1}{2}21​
  3. (C)The area of the region bounded by the ellipse between the lines x=12x = \frac{1}{\sqrt{2}}x=2​1​ and x=1x = 1x=1 is 142(π−2)\frac{1}{4\sqrt{2}}(\pi - 2)42​1​(π−2)
  4. (D)The area of the region bounded by the ellipse between the lines x=12x = \frac{1}{\sqrt{2}}x=2​1​ and x=1x = 1x=1 is 116(π−2)\frac{1}{16}(\pi - 2)161​(π−2)

Correct answer: (A), (C)

Step-by-step solution →
Q110·MathematicsSingle correctJEE Advanced 2017
Answer by appropriately matching the information given in the three columns of the following table. Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively. The tangent to a suitable conic (Column 1) at (3,12)\left(\sqrt{3}, \frac{1}{2}\right)(3​,21​) is found to be 3x+2y=4\sqrt{3}x + 2y = 43​x+2y=4, then which of the following options is the only CORRECT combination ?
Column 1Column 2Column 3
(I) x2+y2=a2x^{2} + y^{2} = a^{2}x2+y2=a2(i) my=m2x+amy = m^{2}x + amy=m2x+a(P) (am2,2am)\left(\frac{a}{m^{2}}, \frac{2a}{m}\right)(m2a​,m2a​)
(II) x2+a2y2=a2x^{2} + a^{2}y^{2} = a^{2}x2+a2y2=a2(ii) y=mx+am2+1y = mx + a\sqrt{m^{2} + 1}y=mx+am2+1​(Q) (−mam2+1,am2+1)\left(\frac{-ma}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}}\right)(m2+1​−ma​,m2+1​a​)
(III) y2=4axy^{2} = 4axy2=4ax(iii) y=mx+a2m2−1y = mx + \sqrt{a^{2}m^{2} - 1}y=mx+a2m2−1​(R) (−a2ma2m2+1,1a2m2+1)\left(\frac{-a^{2}m}{\sqrt{a^{2}m^{2} + 1}}, \frac{1}{\sqrt{a^{2}m^{2} + 1}}\right)(a2m2+1​−a2m​,a2m2+1​1​)
(IV) x2−a2y2=a2x^{2} - a^{2}y^{2} = a^{2}x2−a2y2=a2(iv) y=mx+a2m2+1y = mx + \sqrt{a^{2}m^{2} + 1}y=mx+a2m2+1​(S) (−a2ma2m2−1,−1a2m2−1)\left(\frac{-a^{2}m}{\sqrt{a^{2}m^{2} - 1}}, \frac{-1}{\sqrt{a^{2}m^{2} - 1}}\right)(a2m2−1​−a2m​,a2m2−1​−1​)
  1. (A)(II) (iii) (R)
  2. (B)(IV) (iv) (S)
  3. (C)(IV) (iii) (S)
  4. (D)(II) (iv) (R)

Correct answer: (D)

Step-by-step solution →
Q111·MathematicsSingle correctJEE Advanced 2016
Let F1(x1,0)F_{1}(x_{1}, 0)F1​(x1​,0) and F2(x2,0)F_{2}(x_{2}, 0)F2​(x2​,0), for x1<0x_{1} < 0x1​<0 and x2>0x_{2} > 0x2​>0, be the foci of the ellipse x29+y28=1\frac{x^{2}}{9} + \frac{y^{2}}{8} = 19x2​+8y2​=1. Suppose a parabola having vertex at the origin and focus at F2F_{2}F2​ intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. The orthocentre of the triangle F1MNF_{1}MNF1​MN is
  1. (A)(−910,0)\left( -\frac{9}{10}, 0 \right)(−109​,0)
  2. (B)(23,0)\left( \frac{2}{3}, 0 \right)(32​,0)
  3. (C)(910,0)\left( \frac{9}{10}, 0 \right)(109​,0)
  4. (D)(23,6)\left( \frac{2}{3}, \sqrt{6} \right)(32​,6​)

Correct answer: (A)

Step-by-step solution →
Q112·MathematicsSingle correctJEE Advanced 2016
Let F1(x1,0)F_{1}(x_{1}, 0)F1​(x1​,0) and F2(x2,0)F_{2}(x_{2}, 0)F2​(x2​,0), for x1<0x_{1} < 0x1​<0 and x2>0x_{2} > 0x2​>0, be the foci of the ellipse x29+y28=1\frac{x^{2}}{9} + \frac{y^{2}}{8} = 19x2​+8y2​=1. Suppose a parabola having vertex at the origin and focus at F2F_{2}F2​ intersects the ellipse at point M in the first quadrant and at point N in the fourth quadrant. If the tangents to the ellipse at M and N meet at R and the normal to the parabola at M meets the x-axis at Q, then the ratio of area of the triangle MQR to area of the quadrilateral MF1NF2MF_{1}NF_{2}MF1​NF2​ is
  1. (A)3 : 4
  2. (B)4 : 5
  3. (C)5 : 8
  4. (D)2 : 3

Correct answer: (C)

Step-by-step solution →
Q113·MathematicsMultiple correctJEE Advanced 2015
Let E1E_{1}E1​ and E2E_{2}E2​ be two ellipses whose centers are at the origin. The major axes of E1E_{1}E1​ and E2E_{2}E2​ lie along the x-axis and the y-axis, respectively. Let SSS be the circle x2+(y−1)2=2x^{2} + (y - 1)^{2} = 2x2+(y−1)2=2. The straight line x+y=3x + y = 3x+y=3 touches the curves SSS, E1E_{1}E1​ ad E2E_{2}E2​ at PPP, QQQ and RRR, respectively. Suppose that PQ=PR=223PQ = PR = \dfrac{2\sqrt{2}}{3}PQ=PR=322​​. If e1e_{1}e1​ and e2e_{2}e2​ are the eccentricities of E1E_{1}E1​ and E2E_{2}E2​, respectively, then the correct expression(s) is(are)
  1. (A)e12+e22=4340e_{1}^{2} + e_{2}^{2} = \dfrac{43}{40}e12​+e22​=4043​
  2. (B)e1e2=7210e_{1}e_{2} = \dfrac{\sqrt{7}}{2\sqrt{10}}e1​e2​=210​7​​
  3. (C)∣e12−e22∣=58\left|e_{1}^{2} - e_{2}^{2}\right| = \dfrac{5}{8}​e12​−e22​​=85​
  4. (D)e1e2=34e_{1}e_{2} = \dfrac{\sqrt{3}}{4}e1​e2​=43​​

Correct answer: (A), (B)

Step-by-step solution →
Q114·MathematicsIntegerJEE Advanced 2013
A vertical line passing through the point (h,0)(h,0)(h,0) intersects the ellipse x24+y23=1\frac{x^2}{4}+\frac{y^2}{3}=14x2​+3y2​=1 at the points P and Q. Let the tangents to the ellipse at P and Q meet at the point R. If Δ(h)\Delta(h)Δ(h) = area of the triangle PQR, Δ1=max⁡1/2≤h≤1Δ(h)\Delta_1=\max_{1/2\leq h\leq 1}\Delta(h)Δ1​=max1/2≤h≤1​Δ(h) and Δ2=min⁡1/2≤h≤1Δ(h)\Delta_2=\min_{1/2\leq h\leq 1}\Delta(h)Δ2​=min1/2≤h≤1​Δ(h), then 85Δ1−8Δ2=\frac{8}{\sqrt{5}}\Delta_1-8\Delta_2=5​8​Δ1​−8Δ2​= ________

Correct answer: 9

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Ellipse — frequently asked

How many questions from Ellipse appear in JEE?

Ellipse has appeared in 103 of the last 186 JEE Main and JEE Advanced papers — about 55% of them — contributing 114 questions in total across those papers.

Is Ellipse an important chapter for JEE?

Judged by how often it is actually tested, it appears in roughly 55% of papers. Chapters above about 50% are effectively guaranteed to show up every session, so they repay thorough preparation; lower-frequency chapters are better treated as targeted revision.

Where do these Ellipse questions come from?

Every question is from an official JEE Main or JEE Advanced paper, transcribed from the original paper and tagged to this chapter. Answers follow the official answer key.

Other Mathematics chapters

  • Three Dimensional Geometry 344
  • Matrices and Determinants 342
  • Sets, Relations and Functions 313
  • Sequence and Series 287
  • Definite Integration 267
  • Vector Algebra 245
  • Differential Equations 240
  • Probability 226

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