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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 211-240 of 594 questions

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211 Easy
211 of 594
Application +4 | -1
In the ellipse, the minor axis is 8 and eccentricity is 5/3\sqrt{5} / 3. Then major axis is
View answer and explanation
Correct option: B
12
Option B (12) is correct because the major axis is calculated using the eccentricity and minor axis of the ellipse.
212 Easy
212 of 594
Application +4 | -1
If α\alpha and β\beta are eccentric angles of the ends of aa focal chord of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, then tan⁡α2tan⁡β2\tan \frac{\alpha}{2} \tan \frac{\beta}{2} is equal to
View answer and explanation
Correct option: B
e−1e+1\frac{e-1}{e+1}
Option B (e−1e+1\frac{e-1}{e+1}) is correct because it represents the relationship between the tangents of half angles and the eccentricity.
213 Easy
213 of 594
Application +4 | -1
The normal at the point (3,23)(3,2 \sqrt{3}) on the ellipse x236+y216=1\frac{x^2}{36}+\frac{y^2}{16}=1 meets the major axis of the ellipse at (α,0)(\alpha, 0), the value of α\alpha is:
View answer and explanation
Correct option: A
53\frac{5}{3}
Option A (53\frac{5}{3}) is correct because substituting y=0 into the equation of the normal to the ellipse at the given point yields x = 5/3.
214 Medium
214 of 594
Application +4 | -1
Tangent are drawn from any point on the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 to the circle x2+y2=r2x^2+y^2=r^2, then the chord of contact is a tangent to the ellipse, is
View answer and explanation
Correct option: B
a2x2+b2y2=r4a^2 x^2+b^2 y^2=r^4
Option B (a2x2+b2y2=r4a^2 x^2+b^2 y^2=r^4) is correct because it represents the condition for the chord of contact to be tangent to another ellipse.
215 Easy
215 of 594
Application +4 | -1
In an ellipse length of minor axis is 8 and eccentricity is 53\frac{{\sqrt 5 }}{3}. The length of major axis is
View answer and explanation
Correct option: B
12
Option B (12) is correct because the length of the major axis is calculated using the relationship between the minor axis, eccentricity, and major axis of an ellipse.
216 Easy
216 of 594
Application +4 | -1
Locus of the point which divides double ordinate of the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 in the ratio 1:21:2 internally, is
View answer and explanation
Correct option: B
x2a2+9y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{9{y^2}}}{{{b^2}}} = 1
Option B (x2a2+9y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{9{y^2}}}{{{b^2}}} = 1) is correct because it represents the locus of the point dividing the double ordinate in the given ratio.
217 Medium
217 of 594
Analyzing +4 | -1
Three points A,B,CA,B,C are taken on the ellipse x2a2+y2b2=1\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 with eccentric angles θ,θ+α and θ+2α\mathrm{\theta },\mathrm{\theta } + \mathrm{\alpha }\,\mathrm{and }\,\mathrm{\theta } + 2\mathrm{\alpha }, then
View answer and explanation
Correct option: D
The maximum value of area is 334ab\frac{{3\sqrt 3 }}{4}ab
Option D (The maximum value of area is 334ab\frac{{3\sqrt 3 }}{4}ab) is correct because the area of the triangle formed by the points on the ellipse is maximized when cos⁡α=−12\cos \alpha = -\frac{1}{2}, resulting in the maximum area of 334ab\frac{{3\sqrt 3 }}{4}ab.
218 Medium
218 of 594
Application +4 | -1
Equation of the latusrectum of the ellipse 9x2+4y2−18x−8y−23=0  9{x^2} + 4{y^2} - 18x - 8y - 23 = 0\; are
View answer and explanation
Correct option: C
y=1±5y = 1 \pm \sqrt 5
Option C (y=1±5y = 1 \pm \sqrt 5) is correct because these are the equations of the latus rectum of the ellipse after rewriting it in standard form and calculating the eccentricity.
219 Medium
219 of 594
Application +4 | -1
The equation of ellipse, whose distance between the foci is equal to 8 and the distance between the directrices is 18 , is
View answer and explanation
Correct option: D
5x2+9y2=1805 x^2+9 y^2=180
Option D (5x2+9y2=1805 x^2+9 y^2=180) is correct because it satisfies the given conditions for the distance between foci and directrices.
220 Medium
220 of 594
Application +4 | -1
An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the necessary length of the string and the distance between the pins respectively in cms. Are
View answer and explanation
Correct option: D
6+25,256 + 2\sqrt 5 ,2\sqrt 5
Option D (6+25,256 + 2\sqrt 5 ,2\sqrt 5) is correct because the length of the string is the sum of the major axis and the distance between the foci, and the distance between the pins is the distance between the foci.
221 Medium
221 of 594
Application +4 | -1
Let x2a2+y2b2=1(a>b)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b) be a given ellipse, length of whose latusrectum 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, then a2+b2a^2+b^2 is equal to
View answer and explanation
Correct option: B
126
Option B (126) is correct because the values of a and b are calculated using the latus rectum and eccentricity, and then a2+b2a^2 + b^2 is computed.
222 Easy
222 of 594
Application +4 | -1
The vertices of the ellipse 9x2+4y2−18x−27=09 x^2+4 y^2-18 x-27=0, are
View answer and explanation
Correct option: B
(1,±3)(1, \pm 3)
Option B ((1,±3)(1, \pm 3)) is correct because it represents the vertices of the ellipse after shifting the origin.
223 Medium
223 of 594
Understanding +4 | -1
The product of the perpendiculars from the foci of the ellipse x2144+y2100=1\frac{x^2}{144}+\frac{y^2}{100}=1 on any tangent is:
View answer and explanation
Correct option: B
100
Option B (100) is correct because the product of the perpendiculars from the foci of an ellipse to any tangent is equal to the square of the semi-minor axis.
224 Easy
224 of 594
Application +4 | -1
A focus of an ellipse is at the origin. The directrix is the line x=4x = 4 and the eccentricity is 12\frac{1}{2}, then length of semi major axis is
View answer and explanation
Correct option: B
83\frac{8}{3}
Option B (83\frac{8}{3}) is correct because it is calculated using the relationship between the distance from the center to the directrix and the eccentricity of the ellipse.
225 Easy
225 of 594
Application +4 | -1
All ellipse has its centre at (1,−1)\left( {1, - 1} \right) and semi-major axis =8= 8 and it passes through the point (1,3)\left( {1,3} \right). The equation of the ellipse is
View answer and explanation
Correct option: B
(x−1)264+(y+1)216=1\frac{{{{\left( {x - 1} \right)}^2}}}{{64}} + \frac{{{{\left( {y + 1} \right)}^2}}}{{16}} = 1
Option B ((x−1)264+(y+1)216=1\frac{{{{\left( {x - 1} \right)}^2}}}{{64}} + \frac{{{{\left( {y + 1} \right)}^2}}}{{16}} = 1) is correct because it satisfies the given conditions: center at (1, -1), semi-major axis of 8, and passing through the point (1, 3).
226 Medium
226 of 594
Analyzing +4 | -1
If a triangle is inscribed in an ellipse and two of its sides are parallel to given straight lines, then envelope of the third side is
View answer and explanation
Correct option: D
An ellipse
Option D (An ellipse) is correct because the envelope of the third side of the triangle is an ellipse when two sides are parallel to given straight lines.
227 Easy
227 of 594
Application +4 | -1
The equation of the ellipse whose foci are (± 2,0)\left( { \pm \,2,0} \right) and eccentricity 12\frac{1}{2} is
View answer and explanation
Correct option: B
x216+y212=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{12}} = 1
Option B (x216+y212=1\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{12}} = 1) is correct because it satisfies the conditions for the ellipse with given foci and eccentricity.
228 Easy
228 of 594
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.
229 Easy
229 of 594
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.
230 Easy
230 of 594
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.
231 Medium
231 of 594
Application +4 | -1
Let E\mathrm{E} be the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=1 and CC be the circle x2+y2=9x^2+y^2=9. Let PP and QQ be the points (1,2)(1,2) and (2,1)(2,1), respectively. Then
View answer and explanation
Correct option: D
PP lies inside CC but outside EE
Option D (P lies inside C but outside E) is correct because substituting the coordinates of P and Q into the equations of the ellipse and circle determines their positions relative to the curves.
232 Easy
232 of 594
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.
233 Medium
233 of 594
Application +4 | -1
The tangent at the point P(42,32)P\left(\frac{4}{\sqrt{2}}, \frac{3}{\sqrt{2}}\right) to the ellipse 9x2+16y2=1449 x^2+16 y^2=144 meets the axis of xx at AA and the axis of yy at BB. If CC is the centre of the ellipse, then area of the ΔABC\Delta A B C is (in sq. units)
View answer and explanation
Correct option: A
12
Option B (12) is correct because it is the area of the triangle formed by the tangent to the ellipse and the coordinate axes.
234 Easy
234 of 594
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.
235 Easy
235 of 594
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.
236 Easy
236 of 594
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.
237 Easy
237 of 594
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.
238 Easy
238 of 594
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) ?
View answer and explanation
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.
239 Easy
239 of 594
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.
240 Easy
240 of 594
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.
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