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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Mixed difficulty

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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 121-150 of 594 questions

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121 Easy
121 of 594
Analyzing +4 | -1
If the line x+2y+4=0x+2 y+4=0 cutting the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 in points whose eccentric angles are 30∘30^{\circ} and 60∘60^{\circ} subtends a right angle at the orgin, then its equation is
View answer and explanation
Correct option: B
x216+y24=1\frac{x^2}{16}+\frac{y^2}{4}=1
Option B (x216+y24=1\frac{{x^2}}{{16}}+\frac{{y^2}}{{4}}=1) is correct because it satisfies the condition that the line intersects the ellipse at points with eccentric angles 30° and 60° and subtends a right angle at the origin.
122 Easy
122 of 594
Application +4 | -1
If the foci and vertices of an ellipse be (±1,0)\left( { \pm 1,0} \right) and (±2,0)\left( { \pm 2,0} \right) then the minor axis of the ellipse is
View answer and explanation
Correct option: D
232\sqrt 3
Option D (232\sqrt{3}) is correct because it represents the length of the minor axis calculated using the relationship between the semi-major axis, eccentricity, and semi-minor axis of the ellipse.
123 Easy
123 of 594
Application +4 | -1
The locus of the point of intersection of the perpendicular tangents to the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1 is
View answer and explanation
Correct option: C
x2+y2=13x^2+y^2=13
Option C (x2+y2=13x^2+y^2=13) is correct because the locus of the point of intersection of perpendicular tangents to an ellipse is the director circle, given by x2+y2=a2+b2x^2 + y^2 = a^2 + b^2.
124 Easy
124 of 594
Application +4 | -1
The foci of an ellipse are (±2,0)( \pm 2,0) and its eccentricity is 12\frac{1}{2} then the equation of ellipse is x2a2+y212=1\frac{x^2}{a^2}+\frac{y^2}{12}=1. The value of 'a{a}' is
View answer and explanation
Correct option: A
4
Option A (4) is correct because the value of 'a' in the ellipse's equation is 4, derived from the foci and eccentricity.
125 Medium
125 of 594
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.
126 Medium
126 of 594
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.
127 Easy
127 of 594
Application +4 | -1
The value of aa for the ellipse x2a2+y2b2=1(a>b)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a > b), if the extremities of the latusrectum of the ellipse having positive ordinate lies on the parabola x2=−2(y−2)x^2=-2(y-2), is
View answer and explanation
Correct option: A
2
Option A (2) is correct because it satisfies the condition that the extremities of the latus rectum lie on the given parabola.
128 Easy
128 of 594
Application +4 | -1
The line x=at2x=a t^2 meets the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 in the real points iff
View answer and explanation
Correct option: B
∣t∣≤1|t| \leq 1
Option B (∣t∣≤1|t| \leq 1) is correct because for the line to intersect the ellipse in real points, the condition 1−t4≥01-t^4 \geq 0 must be satisfied, which implies ∣t∣≤1|t| \leq 1.
129 Easy
129 of 594
Application +4 | -1
The latusrectum of the ellipse 9x2+15y2=1449{x^2} + 15{y^2} = 144 is
View answer and explanation
Correct option: D
92\frac{9}{2}
Option D (92\frac{9}{2}) is correct because the latus rectum of the ellipse is calculated using the formula 2b2a\frac{{2b^2}}{a} after rewriting the equation in standard form.
130 Easy
130 of 594
Application +4 | -1
For 0<θ<π/20 < \theta < \pi / 2, if the eccentricity of the hyperbola x2−y2cosec⁡2θ=5x^2-y^2 \operatorname{cosec}^2 \theta=5 is 7\sqrt{7} times eccentricity of the ellipse x2cosec⁡2θ+y2=5x^2 \operatorname{cosec}^2 \theta+y^2=5, then the value of θ\theta is:
View answer and explanation
Correct option: C
π3\frac{\pi}{3}
Option C (π3\frac{\pi}{3}) is correct because it is the value of θ\theta that satisfies the given relationship between the eccentricities of the hyperbola and the ellipse.
131 Easy
131 of 594
Application +4 | -1
The equation of the normal at the point P(2,3)P\left( {2,3} \right) on the ellipse 9x2+16y2=1809{x^2} + 16{y^2} = 180 is
View answer and explanation
Correct option: B
3y−8x+7=03y - 8x + 7 = 0
Option B (3y−8x+7=03y - 8x + 7 = 0) is correct because substituting the point into the general normal equation and simplifying yields this result.
132 Medium
132 of 594
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.
133 Easy
133 of 594
Application +4 | -1
The eccentricity of the conic 4x2+16y2−24x−32y=14{x^2} + 16{y^2} - 24x - 32y = 1 is
View answer and explanation
Correct option: C
32\frac{{\sqrt 3 }}{2}
Option C (32\frac{{\sqrt 3 }}{2}) is correct because it is the calculated eccentricity of the given ellipse after completing the square and putting it in standard form.
134 Easy
134 of 594
Application +4 | -1
Tangent is drawn to the ellipse x227+y2=1\frac{x^2}{27}+y^2=1 at (33cos⁡θ,sin⁡θ)(3 \sqrt{3} \cos \theta, \sin \theta) where θ∈(0,π/2)\theta \in(0, \pi / 2). Then the value of θ\theta such that sum of intercepts on axes made by the tangent is minimum, is
View answer and explanation
Correct option: B
π/6\pi / 6
Option B (π6\frac{{\pi}}{{6}}) is correct because it minimizes the sum of the intercepts on the axes made by the tangent.
135 Easy
135 of 594
Application +4 | -1
The ellipse E1:x29+y24=1E_1: \frac{x^2}{9}+\frac{y^2}{4}=1 is inscribed in a rectangle RR whose sides are parallel to the coordinates axes. Another ellipse E2E_2 passing through the point (0,4)(0,4) circumscribing the rectangle RR. The eccentricity of the ellipse E2E_2, is
View answer and explanation
Correct option: C
12\frac{1}{2}
Option C (12\frac{1}{2}) is correct because the eccentricity of the ellipse E2E_2 is calculated using the given information about the ellipses and the rectangle.
136 Medium
136 of 594
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.
137 Medium
137 of 594
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.
138 Easy
138 of 594
Application +4 | -1
Locus of all such points so that sum of its distances from (2,−3)(2,-3) and (2,5)(2,5) is always 10, is
View answer and explanation
Correct option: D
(x−2)29+(y−1)225=1\frac{(x-2)^2}{9}+\frac{(y-1)^2}{25}=1
Option D ((x−2)29+(y−1)225=1\frac{{(x-2)^2}}{{9}}+\frac{{(y-1)^2}}{{25}}=1) is correct because the locus is an ellipse with foci (2,-3) and (2,5), major axis 10, and center (2,1).
139 Easy
139 of 594
Analyzing +4 | -1
From a point on the axis of xx common tangents are drawn to the parabola y2=4xy^2=4 x and the ellipse x2a2+y2b2=1(a>b>0)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0). If these tangents form an equilateral triangle with their chord of contact w.r.t parabola, then set of exhaustive values of aa is
View answer and explanation
Correct option: C
(32,3)\left(\frac{3}{2}, 3\right)
Option C ((32,3)\left(\frac{{3}}{{2}}, 3\right)) is correct because the condition for the tangents to form an equilateral triangle leads to a range of values for the semi-major axis 'a' of the ellipse.
140 Easy
140 of 594
Application +4 | -1
The eccentricity of the ellipse, if the distance between the foci is equal to the length of the latusrectum, is
View answer and explanation
Correct option: A
5−12\frac{\sqrt{5}-1}{2}
Option A (5−12\frac{\sqrt{5}-1}{2}) is correct because it is the eccentricity calculated from the given condition that the distance between the foci equals the length of the latus rectum.
141 Medium
141 of 594
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.
142 Easy
142 of 594
Application +4 | -1
The equation of the chord of x236+y29=1\frac{x^2}{36}+\frac{y^2}{9}=1 which is bisected at (2,1)(2,1), is
View answer and explanation
Correct option: C
x+2y−4=0x+2 y-4=0
Option C (x+2y−4=0x+2 y-4=0) is correct because it represents the equation of the chord bisected at the given point.
143 Easy
143 of 594
Analyzing +4 | -1
If polar of x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is always touching the curve x2b2+y2a2=1\frac{x^2}{b^2}+\frac{y^2}{a^2}=1, then locus of pole is
View answer and explanation
Correct option: B
Ellipse
Option B (Ellipse) is correct because the locus of the pole is an ellipse, derived by applying the tangency condition to the polar and the given curve.
144 Easy
144 of 594
Application +4 | -1
The latus rectum of the ellipse 4x2+9y2−8x−36y+4=04 x^2+9 y^2-8 x-36 y+4=0 is
View answer and explanation
Correct option: A
8/38 / 3
Option A (8 / 3) is correct because the latus rectum is calculated using the formula 2b²/a after rewriting the equation in standard ellipse form.
145 Easy
145 of 594
Application +4 | -1
The eccentricity of the ellipse 25x2+16y2−150x−175=025{x^2} + 16{y^2} - 150x - 175 = 0 is
View answer and explanation
Correct option: D
35\frac{3}{5}
Option D (35\frac{3}{5}) is correct because the eccentricity of the ellipse is calculated using the formula e=1−a2b2e = \sqrt{1 - \frac{a^2}{b^2}} after rewriting the equation in standard form.
146 Medium
146 of 594
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).
147 Easy
147 of 594
Application +4 | -1
If xcos⁡α+ysin⁡α=px\cos \alpha + y\sin \alpha = p is a tangent to the ellipse, then
View answer and explanation
Correct option: C
a2cos⁡2α+b2sin⁡2α=p2{a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha = {p^2}
Option C (a2cos⁡2α+b2sin⁡2α=p2{a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha = {p^2}) is correct because it represents the condition for the given line to be a tangent to the ellipse.
148 Easy
148 of 594
Application +4 | -1
The equations of tangents to the ellipse 3x2+4y2=53{x^2} + 4{y^2} = 5, which are inclined at 300{30^0} to the x−axisx - axis, are
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because the calculated tangent equation does not match any of the provided options.
149 Easy
149 of 594
Application +4 | -1
If xa+yb=2\frac{x}{a} + \frac{y}{b} = \sqrt 2 touches the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 then the eccentric angle of the point of contact is equal to
View answer and explanation
Correct option: C
450{45^0}
The correct answer is C (450{45^0}) because it represents the eccentric angle of the point of contact of the line and the ellipse.
150 Easy
150 of 594
Application +4 | -1
The curve represented by x=3(cos⁡t+sin⁡t),y=4(cos⁡t−sin⁡t)x = 3(\cos t + \sin t),y = 4(\cos t - \sin t) is
View answer and explanation
Correct option: A
Ellipse
Option A (Ellipse) is correct because the given parametric equations can be manipulated to form the equation of an ellipse.
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