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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Easy

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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 151-180 of 392 questions

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151 Easy
151 of 392
Application +4 | -1
Equation of the tangent to the ellipse x24+y23=1\frac{x^2}{4}+\frac{y^2}{3}=1 perpendicular to the line 2x+y+7=02 x+y+7=0, is
View answer and explanation
Correct option: C
x−2y−2=0x-2 y-2=0
Option D (x−2y−5=0x-2 y-5=0) is incorrect. The correct answer should be x−2y−2=0x-2y-2=0 because it satisfies the condition for tangency to the ellipse and is perpendicular to the given line.
152 Easy
152 of 392
Application +4 | -1
Tangents are drawn from the point P(3,4)P(3,4) to the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1 touching the ellipse at points AA and BB. The coordinates of AA and BB, are
View answer and explanation
Correct option: D
(3,0)(3,0) and (−95,85)\left(-\frac{9}{5}, \frac{8}{5}\right)
Option D ((3,0)(3,0) and (−95,85)(-\frac{9}{5}, \frac{8}{5})) is correct because it represents the points of tangency found by solving the chord of contact equation with the ellipse equation.
153 Easy
153 of 392
Application +4 | -1
The eccentricity of the ellipse 9x2+25y2=2259 x^2+25 y^2=225 is
View answer and explanation
Correct option: B
45\frac{4}{5}
Option B (45\frac{{4}}{{5}}) is correct because the eccentricity is calculated using the formula e=1−b2a2e = \sqrt{{1 - \frac{{b^2}}{{a^2}}}} after converting the equation to standard form.
154 Easy
154 of 392
Application +4 | -1
(3x−4y+10)22+(4x+3y−15)23=1\frac{(3 x-4 y+10)^2}{2}+\frac{(4 x+3 y-15)^2}{3}=1 is an ellipse, then find major & minor axes
View answer and explanation
Correct option: D
235\frac{2 \sqrt{3}}{5} and 225\frac{2 \sqrt{2}}{5}
Option D (235\frac{2 \sqrt{3}}{5} and 225\frac{2 \sqrt{2}}{5}) is correct because it correctly identifies the major and minor axes after rewriting the ellipse equation in standard form.
155 Easy
155 of 392
Application +4 | -1
x=4(1+cos⁡θ)x = 4(1 + \cos \mathrm{\theta }) and y=3(1+sin⁡θ)y = 3(1 + \sin \mathrm{\theta }) are the parametric equations of
View answer and explanation
Correct option: D
(x−4)216+(y−3)29=1\frac{{{{\left( {x - 4} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 3} \right)}^2}}}{9} = 1
Option D ((x−4)216+(y−3)29=1\frac{{{{\left( {x - 4} \right)}^2}}}{{16}} + \frac{{{{\left( {y - 3} \right)}^2}}}{9} = 1) is correct because it is the equation of the ellipse derived from the given parametric equations.
156 Easy
156 of 392
Application +4 | -1
S\mathrm{S} and T\mathrm{T} are the foci of the ellipse x2/a2+y2/b2=1\mathrm{x}^2 / \mathrm{a}^2+\mathrm{y}^2 / \mathrm{b}^2=1 and B\mathrm{B} is an end of the minor axis. If STB\mathrm{S T B} is an equilateral triangle, the eccentricity of the ellipse is e\mathrm{e} then find value of 4e4 \mathrm{e}
View answer and explanation
Correct option: A
2
Option A (2) is correct because using the properties of an ellipse and the given equilateral triangle condition, we can derive the eccentricity and find that 4e = 2.
157 Easy
157 of 392
Application +4 | -1
The maximum value of ∣x+y∣|x+y|, where (x,y)(x, y) lies on the ellipse 4x2+y2=44 x^2+y^2=4 is
View answer and explanation
Correct option: B
5\sqrt{5}
Option B (5\sqrt{5}) is correct because it represents the maximum value of ∣x+y∣|x+y| obtained by parameterizing the ellipse and using trigonometric identities.
158 Easy
158 of 392
Application +4 | -1
The equations to the directrices of the ellipse 4(x−3)2+9(y+2)2=1444{\left( {x - 3} \right)^2} + 9{\left( {y + 2} \right)^2} = 144 are
View answer and explanation
Correct option: A
5x−15±185=05x - 15 \pm 18\sqrt 5 = 0
Option A (5x−15±185=05x - 15 \pm 18\sqrt 5 = 0) is correct because it represents the equations of the directrices of the given ellipse.
159 Easy
159 of 392
Application +4 | -1
Which of the following conic has tangent x+3y−23=0x+\sqrt{3} y-2 \sqrt{3}=0 at point (332,12)\left(\frac{3 \sqrt{3}}{2}, \frac{1}{2}\right) ?
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Correct option: C
x2+9y2=9x^2+9 y^2=9
Option C (x2+9y2=9x^2+9 y^2=9) is correct because the point lies on the curve and the tangent at that point matches the given tangent equation.
160 Easy
160 of 392
Application +4 | -1
A circle has the same centre as an ellipse & passes through the focii F1F_1 & F2F_2 of the ellipse, such that the two curves intersect at 4 points. Let ' PP ' be any one of their points of intersection. If the major axis of the ellipse is 1717 & the area of the triangle PF1F2P F_1 F_2 is 30 , then the distance between the focii is :
View answer and explanation
Correct option: C
13
Option C (13) is correct because it uses the properties of the ellipse and the area of the triangle to find the distance between the foci.
161 Easy
161 of 392
Understanding +4 | -1
A point PP moves so that sum of its distances from (−ae,0)\left( {-ae,0} \right) and (ae,0)\left( {ae,0} \right) is 2a2a. Then, the locus of PP is
View answer and explanation
Correct option: D
x2a2−y2a2(1+e2)=1\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{a^2}\left( {1 + {e^2}} \right)}} = 1
Option B (x2a2+y2a2(1−e2)=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{a^2}\left( {1 - {e^2}} \right)}} = 1) is correct because the sum of distances from two points is constant, defining an ellipse.
162 Easy
162 of 392
Application +4 | -1
Let LL′LL' be the latusrectum and S be a focus of the ellipse x2a2+y2b2=1.\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1. If ΔSLL′\Delta SLL' is equilaterial, then the eccentricity of the ellipse is
View answer and explanation
Correct option: B
1/31/\sqrt 3
Option B (1/31/\sqrt 3) is correct because when the triangle SLL' is equilateral, the eccentricity can be derived using the properties of the ellipse and the latus rectum.
163 Easy
163 of 392
Application +4 | -1
Let the equations of two ellipses be
E1:x23+y22=1 and E2:x216+y2b2=1,E_{1}: \frac{x^{2}}{3}+\frac{y^{2}}{2}=1 \text { and } E_{2}: \frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1,

If the product of their eccentricities is 12\frac{1}{2}, then the length of theminor axis of ellipse E2E_{2} is :
View answer and explanation
Correct option: C
4
Option C (4) is correct because the length of the minor axis of ellipse E2E_2 is calculated using the product of eccentricities and the given equation.
164 Easy
164 of 392
Application +4 | -1
The foci of an ellipse are (0,±4)\left( {0, \pm 4} \right) and the equations for the directrices are y=±9y = \pm 9. The equation for the ellipse is
View answer and explanation
Correct option: D
9x2+5y2=1809{x^2} + 5{y^2} = 180
Option D (9x2+5y2=1809{x^2} + 5{y^2} = 180) is correct because it represents the equation of the ellipse with the given foci and directrices.
165 Easy
165 of 392
Application +4 | -1
The locus of midpoints of a focal chord of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is
View answer and explanation
Correct option: A
x2a2+y2b2=±exa\frac{x^2}{a^2}+\frac{y^2}{b^2}=\pm \frac{e x}{a}
Option A (x2a2+y2b2=±exa\frac{{x^2}}{{a^2}}+\frac{{y^2}}{{b^2}}=\pm \frac{{e x}}{{a}}) is correct because the locus of the midpoints of a focal chord is obtained by using the equation of the chord with a given midpoint and the fact that the chord passes through the focus.
166 Easy
166 of 392
Application +4 | -1
The acute angle between the pair of tangents drawn to the ellipse 2x2+3y2=52 x^2+3 y^2=5 from the point (1,3)(1,3) is
View answer and explanation
Correct option: B
tan⁡−1(2475)\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)
Option B is correct because the acute angle between the tangents is found using the formula involving the slopes of the tangents and the given ellipse equation.
167 Easy
167 of 392
Application +4 | -1
The equation of the ellipse passing through (2,1) having e=1/2e = 1/2 is
View answer and explanation
Correct option: A
3x2+4y2=163{x^2} + 4{y^2} = 16
Option A (3x2+4y2=163{x^2} + 4{y^2} = 16) is correct because it satisfies the ellipse equation given the eccentricity and the point (2,1).
168 Easy
168 of 392
Application +4 | -1
If pair of tangents drawn to the ellipse x216+y29=1\frac{x^2}{16}+\frac{y^2}{9}=1 from a point PP, so that angle between the tangents are at right angle, then possible coordinates of the point PP is/are
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because all the points satisfy the equation of the director circle, which is the locus of points from which perpendicular tangents can be drawn.
169 Easy
169 of 392
Application +4 | -1
The equation of the line passing through the centre and bisecting the chord 7x+y−1=07 x+y-1=0 of the ellipse x21+y27=1\frac{x^2}{1}+\frac{y^2}{7}=1 is
View answer and explanation
Correct option: A
x=yx=y
Option A (x=yx=y) is correct because the line passing through the center and bisecting the chord of the ellipse has the same slope as the line connecting the origin to the midpoint of the chord.
170 Easy
170 of 392
Application +4 | -1
The eccentricity 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: D
5/3\sqrt{5} / 3
Option D (5/3\sqrt{5} / 3) is correct because the eccentricity is calculated using the formula e=1−b2a2e = \sqrt{{1 - \frac{{b^2}}{{a^2}}}} after rewriting the equation in standard ellipse form.
171 Easy
171 of 392
Application +4 | -1
If chords of contact of the tangent from two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 are at right angles, then x1x2a2×y1y2b2\frac{x_1x_2}{a^2} \times \frac{y_1y_2}{b^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 must equal -1 due to their perpendicularity.
172 Easy
172 of 392
Remembering +4 | -1
The sum of focal distance of any point on the ellipse with major and minor axes as 2a2a and 2b2b respectively, is equal to
View answer and explanation
Correct option: A
2a2a
Option A (2a2a) is correct because the sum of the focal distances of any point on the ellipse is equal to the length of the major axis, which is 2a.
173 Easy
173 of 392
Application +4 | -1
The eccentricity of conic 4(2y−x−3)2+9(2x+y−1)2=4(2 y-x-3)^2+9(2 x+y-1)^2= 80 is
View answer and explanation
Correct option: B
53\frac{\sqrt{5}}{3}
Option B (53\frac{{\sqrt{{5}}}}{{3}}) is correct because it represents the eccentricity of the ellipse after transforming the given equation into standard form.
174 Easy
174 of 392
Application +4 | -1
The parametric representation of a point of the ellipse whose foci are (3, 0) and (−1,0)\left( { - 1,0} \right) and eccentricity 23\frac{2}{3} is
View answer and explanation
Correct option: D
(1+3cosθ,5sin⁡θ)(1 + 3\mathrm{cos \theta },\sqrt 5 \sin \mathrm{\theta })
The correct answer is D ((1+3cosθ,5sin⁡θ)(1 + 3\mathrm{cos \theta },\sqrt 5 \sin \mathrm{\theta })) because it represents the parametric coordinates of a point on the ellipse with the given foci and eccentricity.
175 Easy
175 of 392
Analyzing +4 | -1
Let E1:x2a2+y2b2=1,  a>b{E_1}:\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,\,\,a > b. Let E2{E_2} be another ellipse such that it touches the end points of major axis of E1{E_1} and the foci of E2{E_2} are the end points of minor axis of E1{E_1}. If E1{E_1} and E2{E_2} have same eccentricities, then its value is
View answer and explanation
Correct option: A
−1+52\frac{{ - 1 + \sqrt 5 }}{2}
Option A (−1+52\frac{{ - 1 + \sqrt 5 }}{2}) is correct because the eccentricity is derived from the geometric relationships between the two ellipses and solving the resulting quadratic equation.
176 Easy
176 of 392
Understanding +4 | -1
If PP is a point on the ellipse x216+y220=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{20}} = 1 whose foci are SS and S′S'. Then PS+PS′PS + PS' is
View answer and explanation
Correct option: B
454\sqrt 5
Option B (4√5) is correct because the sum of the distances from any point on the ellipse to its foci is equal to the length of the major axis, which is 2a = 2√(20) = 4√5.
177 Easy
177 of 392
Analyzing +4 | -1
Locus of the middle points of all chords of x24+y29=1\frac{x^2}{4}+\frac{y^2}{9}=1, which are at a distance of 2 units from the vertex of parabola y2=−8axy^2=-8ax, is
View answer and explanation
Correct option: B
(x24+y29)2=4(x216+y281)\left(\frac{x^2}{4}+\frac{y^2}{9}\right)^2=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)
Option B ((x24+y29)2=4(x216+y281)\left(\frac{x^2}{4}+\frac{y^2}{9}\right)^2=4\left(\frac{x^2}{16}+\frac{y^2}{81}\right)) is correct because it represents the locus of the midpoints of the chords at a given distance from the vertex of the parabola.
178 Easy
178 of 392
Analyzing +4 | -1
If α−β=\alpha-\beta= constant, then the locus of the point of intersection of tangents at P(acos⁡α,bsin⁡α)P(a \cos \alpha, b \sin \alpha) and Q(acos⁡β,bsin⁡β)Q(a \cos \beta, b \sin \beta) to the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, is
View answer and explanation
Correct option: C
An ellipse
Option C (An ellipse) is correct because the locus of the point of intersection of tangents at P and Q to the ellipse is another ellipse.
179 Easy
179 of 392
Application +4 | -1
If f(x)f(x) is a decreasing function, then the set of values of 'kk', for which the major axis of the ellipse x2f(k2+2k+5)+y2f(k+11)=1\frac{x^2}{f\left(k^2+2 k+5\right)}+\frac{y^2}{f(k+11)}=1 is the XX-axis, is
View answer and explanation
Correct option: B
k∈(−3,2)k \in(-3,2)
Option B (k∈(−3,2)k \in(-3,2)) is correct because it represents the interval where the inequality k2+k−6<0k^2 + k - 6 < 0 holds true, ensuring that f(k2+2k+5)>f(k+11)f(k^2+2k+5) > f(k+11) for a decreasing function f(x).
180 Easy
180 of 392
Application +4 | -1
Let PP be the variable point on the ellipse x225+y216=1\frac{x^2}{25}+\frac{y^2}{16}=1 with focii at S&S′S \& S^{\prime}. Then maximum area of △PSS′\triangle \mathrm{PSS}^{\prime}
View answer and explanation
Correct option: B
12
Option B (12) is correct because the maximum area of the triangle is given by abe, where e is the eccentricity.
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