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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Hard

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Practice format and chapter quiz

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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1 Hard

1 of 15

Application +4 | -1

If the tangent at any point

P

on the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

meets the tangents at the ends

B

and

B'

of minor axis at

L

and

L'

respectively, then

BL.B'L' =

View answer and explanation

Correct option: A

{a^2}

Option A (

{a^2}

) is correct because it is the product of the distances from the points where the tangent at any point P on the ellipse meets the tangents at the ends of the minor axis.

2 Hard

2 of 15

Application +4 | -1

The locus of the poles of normal chords of the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1,

View answer and explanation

Correct option: C

\frac{{{a^6}}}{{{x^2}}} + \frac{{{b^6}}}{{{y^2}}} = {\left( {{a^2} - {b^2}} \right)^2}

Option C (

\frac{{{a^6}}}{{{x^2}}} + \frac{{{b^6}}}{{{y^2}}} = {\left( {{a^2} - {b^2}} \right)^2}

) is correct because it represents the locus of the poles of normal chords of the ellipse, derived by finding the equation of the normal and the polar, then eliminating the parameter.

3 Hard

3 of 15

Analyzing +4 | -1

The curve represented by the equation

\frac{x^2}{\sin \sqrt{2}-\cos \sqrt{3}}+\frac{y^2}{\sin \sqrt{3}-\cos \sqrt{2}}=1

View answer and explanation

Correct option: A

An ellipse with foci on

X

-axis

Option A (An ellipse with foci on

X

-axis) is correct because both denominators are positive and the denominator of

x^2

is greater than the denominator of

y^2

4 Hard

4 of 15

Application +4 | -1

Tangent is drawn to the ellipse

\frac{{{x^2}}}{{27}} + {y^2} = 1

\left( {3\sqrt 3 \cos \mathrm{\theta },\sin \mathrm{\theta }} \right)\left[ {\mathrm{where}\,\,\mathrm{\theta } \in \left( {0,\frac{\pi }{2}} \right)} \right]

. Then, the value of

\theta

such that sum of intercepts on axes made by this tangent is minimum, is

View answer and explanation

Correct option: B

\frac{\pi }{6}

Option B (

\frac{\pi }{6}

) is correct because the sum of the intercepts is minimized when

\tan \theta = \frac{1}{\sqrt 3 }

, which occurs at

\theta = \frac{\pi }{6}

5 Hard

5 of 15

Analyzing +4 | -1

Coordinates of the vertices

B

and

C

of a

\Delta A B C

are

(2,0)

and

(8,0)

respectively. The vertex

A

is varing in such

a

way that

4 \tan \left(\frac{B}{2}\right) \cdot \tan \left(\frac{C}{2}\right)=1

. Then, the locus of

A

View answer and explanation

Correct option: A

\frac{(x-5)^2}{25}+\frac{y^2}{16}=1

Option A (

\frac{(x-5)^2}{25}+\frac{y^2}{16}=1

) is correct because the given condition leads to the sum of distances from A to B and C being constant, which defines an ellipse with the given equation.

6 Hard

6 of 15

Analyzing +4 | -1

For the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

, four normals cannot be drawn through a point

(h, 0)

unless (

e

being the eccentricity of the ellipse)

View answer and explanation

Correct option: C

|h| < a e^2

Option D is correct because the condition for four normals to be drawn from a point (h,0) to the ellipse is |h| < ae².

7 Hard

7 of 15

Analyzing +4 | -1

If circumcentre of an equilateral triangle inscribed in

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

, with vertices having eccentric angles

\alpha, \beta, \gamma

, respectively is

\left(x_1, y_1\right)

then

\Sigma \cos \alpha \cos \beta+\Sigma \sin \alpha \sin \beta=

View answer and explanation

Correct option: D

\frac{9 x_1{ }^2}{2 a^2}+\frac{9 y_1^2}{2 b^2}-\frac{3}{2}

Option D (

\frac{9 x_1{ }^2}{2 a^2}+\frac{9 y_1^2}{2 b^2}-\frac{3}{2}

) is correct because it represents the correct relationship between the coordinates of the circumcenter and the eccentric angles of the vertices of the equilateral triangle.

8 Hard

8 of 15

Application +4 | -1

The length of the smallest intercept made by the coordinate axes on any tangent to the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

, is

View answer and explanation

Correct option: A

a + b

Option A (

a + b

) is correct because it represents the minimum length of the intercept made by the coordinate axes on any tangent to the ellipse.

9 Hard

9 of 15

Application +4 | -1

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:

View answer and explanation

Correct option: B

\frac{2}{\sqrt{5}}

The correct answer is B (

\frac{2}{\sqrt{5}}

) because it is derived from the relationship between the minor axis, distance between foci, and eccentricity of an ellipse.

10 Hard

10 of 15

Application +4 | -1

F_1

and

F_2

are the two foci of the ellipse

\frac{x^2}{9}+\frac{y^2}{4}=1

. Let

P

be a point on the ellipse such that

\left|P F_1\right|=2\left|P F_2\right|

where

F_1

and

F_2

are the two foci of the ellipse. The area of

\Delta P F_1 F_2

View answer and explanation

Correct option: B

Option B (4) is correct because the area of the triangle can be calculated using the coordinates of the foci and a point on the ellipse, given the relationship between the distances from the point to the foci.

11 Hard

11 of 15

Analyzing +4 | -1

The locus of the mid-points of chords of the ellipse

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

that touch the circle

{x^2} + {y^2} = {b^2},

View answer and explanation

Correct option: B

{\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}} \right)^2} = {b^2}\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}}} \right)

Option B

{\left( {\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}} \right)^2} = {b^2}\left( {\frac{{{x^2}}}{{{a^4}}} + \frac{{{y^2}}}{{{b^4}}}} \right)

is correct because it represents the locus of the midpoints of chords of the ellipse that touch the given circle.

12 Hard

12 of 15

Application +4 | -1

If the focal distance of an end of the minor axis of an ellipse (referred to its axes as the axes of

x

and

y

respectively) is

k

and the distance between its foci is

2h

, then its equation is

View answer and explanation

Correct option: B

\frac{{{x^2}}}{{{k^2}}} + \frac{{{y^2}}}{{{k^2} - {h^2}}} = 1

Option B (

\frac{{{x^2}}}{{{k^2}}} + \frac{{{y^2}}}{{{k^2} - {h^2}}} = 1

) is correct because it correctly relates the focal distance and the distance between the foci to the equation of the ellipse.

13 Hard

13 of 15

Analyzing +4 | -1

The normals at four points on the ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

meet in the point

(h, k)

. Then, the mean position of the four points is

View answer and explanation

Correct option: D

\left(\frac{a^2 h}{2\left(a^2-b^2\right)}, \frac{b^2 k}{2\left(a^2-b^2\right)}\right)

Option D is correct because the mean position is derived from the properties of normals to the ellipse and the equation of the ellipse itself.

14 Hard

14 of 15

Application +4 | -1

Let

\mathrm{E}

be the ellipse

\frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{4}=1

and

\mathrm{C}

be the circle centered at

(0,0)

with radius 3 . Let

\mathrm{P}

and

\mathrm{Q}

be the points

(1,2)

and

(2,1)

respectively then -

View answer and explanation

Correct option: C

\mathrm{P}

lies inside

\mathrm{C}

but outside

\mathrm{E}

Option C (P lies inside C but outside E) is correct because when the coordinates of P are substituted into the equations of the ellipse and the circle, it satisfies the conditions.

15 Hard

15 of 15

Analyzing +4 | -1

A tangent to the ellipse

{x^2} + 4{y^2} = 4

meets the ellipse

{x^2} + 2{y^2} = 6

P

and

Q

. The angle between the tangent at

P

and

Q

of the ellipse

{x^2} + 2{y^2} = 6

View answer and explanation

Correct option: A

\frac{\pi }{2}

The correct answer is

\frac{\pi }{2}

because the locus of the point of intersection of tangents to the second ellipse is the director circle, implying a right angle between the tangents.

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Key Takeaways

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