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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-60 of 187 questions
31 Medium
31 of 187
Application +4 | -1
Length of major axis of ellipse 9x2+7y2=639{x^2} + 7{y^2} = 63 is
View answer and explanation
Correct option: C
6
Option C (6) is correct because after converting to standard form, the length of the major axis is twice the value of b, where b is the semi-major axis.
32 Medium
32 of 187
Application +4 | -1
If the chords of constant of tangents from two points (x1,y1)\left( {{x_1},{y_1}} \right) and (x2,y2)\left( {{x_2},{y_2}} \right) to the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 are at right angles, then x1x2y1y2\frac{{{x_1}{x_2}}}{{{y_1}{y_2}}} is equal to
View answer and explanation
Correct option: C
−a4b4- \frac{{{a^4}}}{{{b^4}}}
Option C (−a4b4- \frac{{{a^4}}}{{{b^4}}}) is correct because the product of the slopes of the chords of contact being -1 leads to the given relationship between the coordinates of the points.
33 Medium
33 of 187
Application +4 | -1
Equation of a tangent to the ellipse x236+y225=1\frac{x^2}{36}+\frac{y^2}{25}=1, passing through the point where a directrix of the ellipse meets the + ve xx-axis is
View answer and explanation
Correct option: C
6y+11x=366 y+\sqrt{11} x=36
Option C (6y+11x=366 y+\sqrt{11} x=36) is correct because it represents the tangent to the ellipse passing through the point where the directrix meets the positive x-axis.
34 Medium
34 of 187
Application +4 | -1
An ellipse passing through the point (213,4)(2 \sqrt{13}, 4) has its foci at (−4,1)(-4,1) and (4,1).(4,1). Then its eccentricity is
View answer and explanation
Correct option: A
12\frac{1}{2}
Option A (12\frac{{1}}{{2}}) is correct because the eccentricity is calculated using the distances from a point on the ellipse to its foci, resulting in e = 1/2.
35 Medium
35 of 187
Application +4 | -1
The locus of the foot of the perpendicular drawn from the centre to any tangent to the ellipse x2/a2+y2/b2=1x^2 / a^2+y^2 / b^2=1 is
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the locus of the foot of the perpendicular from the center to any tangent of the ellipse is neither a circle, an ellipse, nor a hyperbola.
36 Medium
36 of 187
Application +4 | -1
Let SS,S′S' be the foci and BB′BB' be the minor axis of the ellipse x2a2+y2b2=1.\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1. If ∠BSS′=θ,\angle BSS' = \theta , then the eccentricity ee of the ellipse is equal to
View answer and explanation
Correct option: B
cos⁡θ\cos \theta
The correct answer is cos⁡θ\cos \theta because it represents the eccentricity of the ellipse, derived from the given angle and the properties of the ellipse.
37 Medium
37 of 187
Application +4 | -1
If P(θ)P\left( \theta \right) and Q(π/2+θ)Q\left( {\pi /2 + \theta } \right) are two points on the ellipse x2a2+y2b2=1.\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1. Locus of the mid-point of PQPQ is
View answer and explanation
Correct option: A
x2a2+y2b2=12\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = \frac{1}{2}
Option A (x2a2+y2b2=12\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = \frac{1}{2}) is correct because it represents the locus of the midpoint of PQ, derived using parametric coordinates and trigonometric identities.
38 Medium
38 of 187
Application +4 | -1
If the tangent at P(θ)P(\theta) on the ellipse 16x2+11y2=25616x^2 + 11y^2 = 256 touches the circle x2+y2+2x–15=0,x^2 + y^2 + 2x – 15 = 0, then θ=\theta =
View answer and explanation
Correct option: C
2π3\frac{2\pi}{3}
Option C (2π3\frac{{2\pi}}{{3}}) is correct because it satisfies the condition where the tangent to the ellipse touches the given circle.
39 Medium
39 of 187
Analyzing +4 | -1
From a point P, perpendicular tangents PQP Q and PRP R are drawn to ellipse x2+4y2=4x^2+4 y^2=4. The locus of the circumcentre of ΔPQR\Delta P Q R, is
View answer and explanation
Correct option: B
x2+y2=516(x2+4y2)2x^2+y^2=\frac{5}{16}\left(x^2+4 y^2\right)^2
Option B (11+sin⁡2θ\frac{1}{\sqrt{1+\sin ^2 \theta}}) is correct because it represents the eccentricity of the ellipse when the chord QR subtends a right angle at the center C.
40 Medium
40 of 187
Application +4 | -1
Let y=2xy=2 x be one of the diagonal of a parallelogram inscribed in an ellipse x2a2+y2b2=1;a>b\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ; a>b having eccentricity a2−8a\frac{\sqrt{a^2-8}}{a}. If the maximum area of parallelogram is 16216 \sqrt{2}, then the value of a is
View answer and explanation
Correct option: B
4
Option B (4) is correct because it involves finding the maximum area of a parallelogram inscribed in an ellipse, using the given eccentricity to determine the value of 'a'.
41 Medium
41 of 187
Application +4 | -1
Suppose SS and S′S' are foci of the ellipse x225+y216=1.\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1. If PP is a variable point on the ellipse and if Δ\Delta is area of the triangles PSS′PSS', then the maximum value Δ\Delta is
View answer and explanation
Correct option: B
12
Option B (12) is correct because it represents the maximum area of the triangle formed by the foci and a point on the ellipse.
42 Medium
42 of 187
Analyzing +4 | -1
An ellipse passes through the foci of the hyperbola, 9x2−4y2=369 x^{2}-4 y^{2}=36 and its major and minor axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of eccentricities of the two conics is 12\frac{1}{2}, then which of the following points does not lie on the ellipse?
View answer and explanation
Correct option: C
(1213,32)\left(\frac{1}{2} \sqrt{13}, \frac{\sqrt{3}}{2}\right)
Option C ((1213,32)(\frac{{1}}{{2}} \sqrt{{13}}, \frac{{\sqrt{{3}}}}{{2}})) is correct because this point does not satisfy the equation of the ellipse.
43 Medium
43 of 187
Application +4 | -1
The eccentricity, foci and the length of the latus rectum of the ellipse x2+4y2+8y−2x+1=0x^2+4 y^2+8 y-2 x+1=0 are respectively equal to:
View answer and explanation
Correct option: C
32;(1±3,−1);1\frac{\sqrt{3}}{2} ;(1 \pm \sqrt{3},-1) ; 1
Option C (32;(1±3,−1);1\frac{{\sqrt{3}}}{{2}} ;(1 \pm \sqrt{3},-1) ; 1) is correct because it correctly identifies the eccentricity, foci, and length of the latus rectum after rewriting the ellipse equation in standard form.
44 Medium
44 of 187
Application +4 | -1
If the coordinates of the vertices of an ellipse are (−6,1)\left( { - 6,1} \right) and (4,1)\left( {4,1} \right) and the equation of a focal chord passing through the focus on the right side of the centre is 2x−y−5=0.2x - y - 5 = 0. The equation of the ellipse is
View answer and explanation
Correct option: B
(x+1)225+(y−1)216=1\frac{{{{\left( {x + 1} \right)}^2}}}{{25}} + \frac{{{{\left( {y - 1} \right)}^2}}}{{16}} = 1
Option B ((x+1)225+(y−1)216=1\frac{{{{\left( {x + 1} \right)}^2}}}{{25}} + \frac{{{{\left( {y - 1} \right)}^2}}}{{16}} = 1) is correct because it represents the equation of the ellipse with the given vertices and focal chord.
45 Medium
45 of 187
Application +4 | -1
An ellipse has foci (3,1),(1,1)(3,1),(1,1) and it passes through point (1,3)(1,3). Its eccentricity is equal to
View answer and explanation
Correct option: A
2−1\sqrt{2}-1
Option A 2−1\sqrt{2}-1 is correct because by using the definition of an ellipse and the given foci and point, we can solve for the eccentricity.
46 Medium
46 of 187
Application +4 | -1
If the line lx+my+n=0l x+m y+n=0 cuts the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 in points whose eccentric angles differ by π/2\pi / 2, then
View answer and explanation
Correct option: A
a2l2+b2m2=2n2a^2 l^2+b^2 m^2=2 n^2
Option A (a2l2+b2m2=2n2a^2 l^2+b^2 m^2=2 n^2) is correct because it represents the condition for the line to intersect the ellipse such that the eccentric angles differ by π/2\pi/2.
47 Medium
47 of 187
Analyzing +4 | -1
If a number of ellipses is described having the same major axis but a variable minor axis, such that each one of the tangents at the ends of their latusrectum passes through the fixed point, then the point of concurrency is
View answer and explanation
Correct option: B
(0,1)(0,1)
Option B ((0,1)) is correct because the tangents at the ends of the latus rectum pass through the fixed point (0, 1).
48 Medium
48 of 187
Application +4 | -1
If P1P_{1} and P2P_{2} are two points on the ellipse x24+y2=1\frac{x^{2}}{4}+y^{2}=1 at which the tangents are parallel to the chord joining the points (0,1)(0,1) and (2,0)(2,0), then the distance between P1P_{1} and P2P_{2} is
View answer and explanation
Correct option: D
10\sqrt{10}
Option D (10\sqrt{10}) is correct because it represents the distance between the two points on the ellipse where the tangents are parallel to the given chord.
49 Medium
49 of 187
Analyzing +4 | -1
In the normal at the end of latusrectum of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 with eccentricity ee, passes through one end of the minor axis, then
View answer and explanation
Correct option: B
e2(1+e2)=1{e^2}\left( {1 + {e^2}} \right) = 1
The correct answer is B. By finding the equation of the normal at the end of the latus rectum and using the given condition, the relationship e2(1+e2)=1{e^2}(1 + {e^2}) = 1 is derived.
50 Medium
50 of 187
Application +4 | -1
If A(1,4)A(1,4) and B(3,0)B(3,0) then the maximum area of the triangle △ABC\triangle A B C inscribed inscribed in ellipse 2x2+y2=182 x^2+y^2=18 is
View answer and explanation
Correct option: B
36+63 \sqrt{6}+6
Option B (36+63 \sqrt{6}+6) is correct because it represents the maximum area of the triangle inscribed in the given ellipse, calculated using parametric coordinates and maximizing the area function.
51 Medium
51 of 187
Application +4 | -1
The normal at (2,32)\left(2, \frac{3}{2}\right) to the ellipse, x216+y23=1\frac{x^{2}}{16}+\frac{y^{2}}{3}=1 touches a parabola, whose equation is
View answer and explanation
Correct option: A
y2=−104xy^{2}=-104 x
Option A (y2=−104xy^{2}=-104 x) is correct because the normal to the ellipse at the given point touches the parabola y2=4axy^2 = 4ax when the condition a−mc=0a - mc = 0 is satisfied.
52 Medium
52 of 187
Application +4 | -1
If M1{M_1} and M2{M_2} are the feet of the perpendiculars from the foci S1{S_1} and S2{S_2} of the ellipse x29+y216=1\frac{{{x^2}}}{9} + \frac{{{y^2}}}{{16}} = 1 on the tangent at any point PP on the ellipse, then (S1M1)(S2M2)\left( {{S_1}{M_1}} \right)\left( {{S_2}{M_2}} \right) is equal to
View answer and explanation
Correct option: A
16
Option A (16) is correct because the product of the perpendiculars from the foci to any tangent of the ellipse is equal to the square of the semi-minor axis.
53 Medium
53 of 187
Analyzing +4 | -1
The locus of the point of the chord of contact of tangents from which the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 touches the circle, is
View answer and explanation
Correct option: A
x2a4+y2b4=1c2\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{c^2}
Option A (x2a4+y2b4=1c2\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{c^2}) is correct because it represents the locus of the point from which the chord of contact of tangents to the ellipse touches the circle.
54 Medium
54 of 187
Application +4 | -1
If tangents are drawn to the ellipse x2+2y2=2x^2+2 y^2=2 then locus of mid-points of intercepts made by the tangents between the coordinate axes is
View answer and explanation
Correct option: A
12x2+14y2=1\frac{1}{2 x^2}+\frac{1}{4 y^2}=1
Option A (12x2+14y2=1\frac{{1}}{{2 x^2}}+\frac{{1}}{{4 y^2}}=1) is correct because it represents the locus of the midpoints of the intercepts formed by the tangents to the ellipse.
55 Medium
55 of 187
Application +4 | -1
If normal to ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 at (ae,b2a)\left(a e, \frac{b^2}{a}\right) is passing through (0,−2b)(0,-2 b), then value of eccentricity is
View answer and explanation
Correct option: C
2(2−1)\sqrt{2(\sqrt{2}-1)}
Option C, 2(2−1)\sqrt{2(\sqrt{2}-1)}, is the correct eccentricity value derived by using the equation of the normal to the ellipse and the given point.
56 Medium
56 of 187
Application +4 | -1
If F1F_1 and F2F_2 are the foci of the ellipse 4x2+9y2=364 x^2+9 y^2=36, PP is a point on the ellipse such that PF1:PF2=P F_1: P F_2= 2:12: 1; then area of ΔPF1F2\Delta P F_1 F_2 is:
View answer and explanation
Correct option: D
4 sq. unit
Option D (4 sq. unit) is correct because the area of the triangle is calculated using the properties of the ellipse and the given ratio of distances to the foci.
57 Medium
57 of 187
Application +4 | -1
The equation of the ellipse whose axes are along the co-ordinate axes, vertices are (±5,0)( \pm 5,0) and foci at (±4,0)( \pm 4,0), is x225+y2b2=1\frac{x^2}{25}+\frac{y^2}{b^2}=1. The value of b2{b}^2 is
View answer and explanation
Correct option: A
9
Option A (9) is correct because the value of b² for the given ellipse is 9, calculated using the vertices and foci.
58 Medium
58 of 187
Analyzing +4 | -1
x2f(4a)+y2f(a2−5)=1\frac{x^2}{f(4 a)}+\frac{y^2}{f\left(a^2-5\right)}=1 represents an ellipse with major axis as y−y- axis and ' ff ' is decreasing function, positive for all ' aa ', then a belongs to
View answer and explanation
Correct option: C
(−1,5)(-1,5)
Option C ((-1,5)) is correct because it satisfies the condition for the ellipse to have its major axis along the y-axis, given that f is a decreasing function.
59 Medium
59 of 187
Application +4 | -1
Consider the particle travelling clockwise on the elliptical path x2100+y225=1\frac{x^2}{100}+\frac{y^2}{25}=1. The particle leaves the orbit at the point (−8,3)(-8,3) and travels in a straight line tangent to the ellipse, then the point will the particle cross the YY-axis is
View answer and explanation
Correct option: C
(0,253)\left(0, \frac{25}{3}\right)
Option C ((0, 253\frac{25}{3})) is correct because the tangent line equation is found and then the y-intercept is calculated by setting x=0.
60 Medium
60 of 187
Application +4 | -1
Let P,Q,RP, Q, R be three points on the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 and P′,Q′P^{\prime}, Q^{\prime} and R′R^{\prime} be the corresponding points on the auxiliary circle. Then,
Area of △PQR\triangle P Q R : Area of △P′Q′R′=\triangle P^{\prime} Q^{\prime} R^{\prime}=
View answer and explanation
Correct option: B
b:ab: a
Option B (b:ab: a) is correct because the ratio of the areas of the triangles is equal to the ratio of the semi-minor axis to the semi-major axis.
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