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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 1-30 of 392 questions

1 Easy

1 of 392

Application +4 | -1

Let

P

be a variable point on the ellipse

\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1

with foci at

S

and

S'.

A

be the area of triangle

PSS',

then the maximum value of

A

View answer and explanation

Correct option: B

12 sq. units

Option B (12 sq. units) is correct because the maximum area of the triangle PSS' is half the product of the distance between the foci and the semi-minor axis.

2 Easy

2 of 392

Application +4 | -1

The slope of a common tangent to the ellipse

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

and a concentric circle of radius

r

, is

View answer and explanation

Correct option: B

\sqrt{\frac{r^2-b^2}{a^2-r^2}}

Option B (

\sqrt{\frac{r^2-b^2}{a^2-r^2}}

) is correct because equating the condition for tangency of a line to both the ellipse and the circle yields this slope.

3 Easy

3 of 392

Analyzing +4 | -1

A tangent to the ellipse

\frac{x^2}{25}+\frac{y^2}{16}=1

at any point

P

meet the line

x=0

at a point

Q

. Let

R

be the image of

Q

in the line

y=x

, then circle whose extremities of a diameter are

Q

and

R

passes through a fixed point. The fixed point is

View answer and explanation

Correct option: D

(0,0)

Option D ((0,0)) is correct because the circle formed with diameter QR always passes through the origin, regardless of the point P on the ellipse.

4 Easy

4 of 392

Analyzing +4 | -1

The point at shortest distance from the line

x+y=7

and lying on an ellipse

x^2+2 y^2=6

, has coordinates

View answer and explanation

Correct option: C

(2,1)

Option C ((2,1)) is correct because the tangent to the ellipse at this point is parallel to the given line, indicating the shortest distance.

5 Easy

5 of 392

Application +4 | -1

For the ellipse

\frac{x^2}{64}+\frac{y^2}{28}=1

, the eccentricity is

View answer and explanation

Correct option: A

\frac{3}{4}

Option A (

\frac{{3}}{{4}}

) is correct because the eccentricity is calculated using the formula

e = \sqrt{{1 - \frac{{b^2}}{{a^2}}}}

with the given values of a and b.

6 Easy

6 of 392

Understanding +4 | -1

If equation of the ellipse is

\frac{x^2}{100}+\frac{y^2}{400}=1

, then

I. Vertices of the ellipse are

(0, \pm 20)

II. Foci of the ellipse are

(0, \pm 10 \sqrt{3})

III. Length of major axis is 40

IV. Eccentricity of the ellipse is

\frac{\sqrt{3}}{2}

View answer and explanation

Correct option: D

All are true

Option D is correct because all the statements (I, II, III, and IV) are true regarding the given ellipse equation.

7 Easy

7 of 392

Application +4 | -1

\alpha, \beta

are the eccentric angles of the extremities of a focal chord of the ellipse

x^2 / 16+y^2 / 9=1

, then

\tan (\alpha / 2) \tan (\beta / 2)=

View answer and explanation

Correct option: D

\frac{8 \sqrt{7}-23}{9}

Option D (

\frac{8 \sqrt{7}-23}{9}

) is correct because it represents the value of

\tan(\alpha/2) \tan(\beta/2)

for a focal chord of the ellipse.

8 Easy

8 of 392

Application +4 | -1

The total number of tangents through the points

\left( {3,5} \right)

that can be drawn to the ellipses

3{x^2} + 5{y^2} = 32

and

25{x^2} + 9{y^2} = 450

View answer and explanation

Correct option: C

Option A (3) is correct because the point (3,5) lies outside the first ellipse (2 tangents) and on the second ellipse (1 tangent).

9 Easy

9 of 392

Application +4 | -1

The eccentricity of an ellipse

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

whose latusrectum is half of its major axis is

View answer and explanation

Correct option: A

\frac{1}{{\sqrt 2 }}

Option A (

\frac{1}{{\sqrt 2 }}

) is correct because the eccentricity is derived from the relationship between the latus rectum and the major axis.

10 Easy

10 of 392

Application +4 | -1

The eccentricity of the ellipse

a x^2+b y^2+2 f x

+2 g y+c=0

if axis of ellipse parallel to

X

-axis is

View answer and explanation

Correct option: A

\sqrt{\left(\frac{b-a}{b}\right)}

Option A (

\sqrt{\left(\frac{{b-a}}{{b}}\right)}

) is correct because it is the standard formula for the eccentricity of an ellipse with its major axis parallel to the x-axis.

11 Easy

11 of 392

Application +4 | -1

Equation of a directrix of the ellipse

\frac{x^2}{36}+\frac{y^2}{4}=1

View answer and explanation

Correct option: C

\sqrt{2} x+9=0

Option C (

\sqrt{2} x+9=0

) is correct because it represents the equation of a directrix of the given ellipse.

12 Easy

12 of 392

Application +4 | -1

The centre of the ellipse

4 x^2+9 y^2-16 x-54 y+61=0

View answer and explanation

Correct option: C

(3,2)

Option B ((2,3)) is correct because completing the square allows us to rewrite the equation in standard ellipse form, revealing the center.

13 Easy

13 of 392

Understanding +4 | -1

An ellipse is the set of all points in a plane, the sum of whose
distances from two fixed points in the plane is a constant.

I. The two fixed points are called the foci of the ellipse.

II. The mid point of the line segment joining the foci is
called the centre of the ellipse.

III. The end points of the major axis are called the vertices
of the ellipse.

View answer and explanation

Correct option: D

All are correct

Option D is correct because all the statements (I, II, and III) are correct definitions related to an ellipse.

14 Easy

14 of 392

Application +4 | -1

The eccentricity of the ellipse

x^2+4 y^2+8 y-2 x+1=0

, is

View answer and explanation

Correct option: A

\frac{\sqrt{3}}{2}

Option A (

\frac{\sqrt{3}}{2}

) is correct because the eccentricity is calculated using the formula

e = \sqrt{1 - \frac{b^2}{a^2}}

after rewriting the ellipse equation in standard form.

15 Easy

15 of 392

Application +4 | -1

The distance of the centre of ellipse

{x^2} + 2{y^2} - 2 = 0

to those tangents of the ellipse which are equally inclined from both the axes, is

View answer and explanation

Correct option: D

\frac{{\sqrt 3 }}{2}

Option B (

\sqrt {\frac{3}{2}}

) is incorrect. Option D (

\frac{{\sqrt 3 }}{2}

) is correct because the distance from the center of the ellipse to the tangents equally inclined to both axes is

\frac{{\sqrt 3 }}{2}

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16 Easy

16 of 392

Application +4 | -1

The centre of the ellipse

\frac{{{{\left( {x + y - 2} \right)}^2}}}{9} + \frac{{{{\left( {x - y} \right)}^2}}}{{16}} = 1

View answer and explanation

Correct option: B

\left( {1,1} \right)

The correct answer is B ((1,1)) because it represents the center of the given ellipse.

17 Easy

17 of 392

Application +4 | -1

The eccentricity of the ellipse

\frac{(x-1)^2}{9}+\frac{(y+1)^2}{25}=1

View answer and explanation

Correct option: A

4 / 5

Option A (4 / 5) is correct because the eccentricity is found using the formula

e = \sqrt{{1 - \frac{{a^2}}{{b^2}}}}

where a² = 9 and b² = 25.

18 Easy

18 of 392

Application +4 | -1

The distance between the directrices of the ellipse

\frac{x^2}{36}+\frac{y^2}{20}=1

View answer and explanation

Correct option: C

Option C (

1 / \sqrt{3}

) is correct because the eccentricity is calculated using the given condition

\frac{2a}{e} = 6ae

19 Easy

19 of 392

Application +4 | -1

Angle between tangents drawn from the point

\left( {5,4} \right)

to the ellipse

\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{16}} = 1

, is

View answer and explanation

Correct option: B

{90^0}

Option C (

(3,4),(-13,4)

) is correct because these are the coordinates of the endpoints of the latus rectum of the given parabola after transforming it to standard form.

20 Easy

20 of 392

Application +4 | -1

The length of major and minor axis of an ellipse are 10 and 8 respectively and its major axis along

y - axis

the equation of the ellipse referred to its centre as origin is

View answer and explanation

Correct option: B

\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{25}} = 1

Option B (

\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{25}} = 1

) is correct because it represents the equation of the ellipse with the given major and minor axes along the y-axis.

21 Easy

21 of 392

Application +4 | -1

a=\left[t^2-3 t+4\right]

and

b=[3+5 t]

, where [.] denotes the greatest integer function, then the latusrectum of the ellipse

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

t=\frac{3}{2}

View answer and explanation

Correct option: C

\frac{1}{5}

Option D (

\frac{1}{10}

) is correct because the latus rectum length is calculated using the greatest integer function and the given ellipse equation at t=3/2.

22 Easy

22 of 392

Application +4 | -1

The foci of the Basic Terms of Conics

25{x^2} + 16{y^2} - 150x = 175

are

View answer and explanation

Correct option: C

\left( {3, \pm 3} \right)

The correct answer is C ((3, ±3)) because completing the square and transforming the equation reveals a vertical ellipse centered at (3,0), and calculating the foci yields (3, ±3).

23 Easy

23 of 392

Understanding +4 | -1

The equation

x=a \cos \theta, y=b \sin \theta(a>b)

represents a conic section whose eccentricity

e

is given by

View answer and explanation

Correct option: C

e^2=\frac{a^2-b^2}{a^2}

Option C (

e^2=\frac{a^2-b^2}{a^2}

) is correct because the equation represents an ellipse, and the eccentricity of an ellipse is given by

e = \sqrt{1 - \frac{b^2}{a^2}}

, so

e^2 = \frac{a^2 - b^2}{a^2}

24 Easy

24 of 392

Application +4 | -1

The locus of pole of tangents to the ellipse

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

with respect to the parabola

y^2=4 a x

, is

View answer and explanation

Correct option: C

b^2 y^2=4 a^2\left(x^2-a^2\right)

Option C (

b^2 y^2 = 4 a^2 (x^2 - a^2)

) is correct because the locus of the pole is found by equating the polar equation to the tangent equation and using the condition for tangency.

25 Easy

25 of 392

Application +4 | -1

If the normal at any point

P

on the ellipse

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

, meets the axes in

G

and

g

respectively, then

P G: P g=

View answer and explanation

Correct option: C

b^2: a^2

Option C (

b^2: a^2

) is correct because it represents the ratio of the distances PG and Pg after substituting the coordinates and simplifying.

26 Easy

26 of 392

Application +4 | -1

2y = x

and

3y + 4x = 0

are the equations of a pair of conjugate diameters of an ellipse, then the eccentricity of the ellipse is

View answer and explanation

Correct option: C

\sqrt {\frac{1}{3}}

Option C (

\sqrt {\frac{1}{3}}

) is correct because it is derived from the relationship between the slopes of conjugate diameters and the eccentricity of the ellipse.

27 Easy

27 of 392

Understanding +4 | -1

The equation

\frac{x^2}{10-a}+\frac{y^2}{4-a}=1

represents an ellipse, if

View answer and explanation

Correct option: A

a < 4

Option A (

a < 4

) is correct because for the equation to represent an ellipse, both denominators must be positive, requiring

a

to be less than 4.

28 Easy

28 of 392

Application +4 | -1

e

is eccentricity of ellipse

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

and

e'

is eccentricity of

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\left( {a < b} \right),

then

View answer and explanation

Correct option: C

\frac{1}{{{e^2}}} + \frac{1}{{{{\left( {e'} \right)}^2}}} = 1

Option C (

\frac{1}{{{e^2}}} + \frac{1}{{{{\left( {e'} \right)}^2}}} = 1

) is correct because it represents the relationship between the eccentricities of an ellipse and its auxiliary ellipse.

29 Easy

29 of 392

Application +4 | -1

The equation of the parabola whose vertex is at the centre of the ellipse

\frac{x^2}{25}+\frac{y^2}{16}=1

and the focus coincide with the focus of the ellipse on the positive side of the major axis of the ellipse is

View answer and explanation

Correct option: D

y^2=12 x

Option D (

y^2=12 x

) is correct because the focus of the ellipse is (3,0), which becomes the focus of the parabola, and the vertex is (0,0).

30 Easy

30 of 392

Application +4 | -1

The eccentricity of an ellipse whose centre is at the origin is

\frac{1}{2}

. If one directrix is

x=-4

, then the equation of the normal to it at

(1,3 / 2)

, is

View answer and explanation

Correct option: B

4 x-2 y=1

Option B (

4 x-2 y=1

) is correct because it is the equation of the normal to the ellipse at the given point.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 392 single-correct-answer MCQs on ELLIPSE.
✓ Try each question before opening its answer. Read the explanation where one is available.
✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.

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