If the line $x\cos \alpha + y\sin {\rm{a}} = p$ be normal to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, then

${p^2}({a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha ) = {a^2} - {b^2}$

${p^2}({a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha ) = {\left( {{a^2} - {b^2}} \right)^2}$

${p^2}({a^2}{\sec ^2}\alpha + {b^2}{\rm{cose}}{{\rm{c}}^2}\alpha ) = {a^2} - {b^2}$

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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 31-60 of 392 questions

31 Easy

31 of 392

Application +4 | -1

In an ellipse, the distance between its foci is 6 and minor axis is 8 . The eccentricity, is

View answer and explanation

Correct option: D

3 / 5

Option D (

3 / 5

) is correct because using the given information about the foci and minor axis, we can calculate the eccentricity as

e = 3/5

32 Easy

32 of 392

Understanding +4 | -1

S

and

S'

are two foci of an ellipse

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

and

P\left( {{x_1},{y_1}} \right)

a point on it, then

SP + S'P

is equal to

View answer and explanation

Correct option: B

2b

Option B (2b) is correct because the sum of the focal distances from any point on an ellipse is equal to the length of the major axis (2b) when a < b.

33 Easy

33 of 392

Application +4 | -1

If the eccentricity of an ellipse be

1 / \sqrt{2}

, then its latus rectum is equal to its

View answer and explanation

Correct option: D

Semi-major axis

Option C (18) is correct because the distance between the directrices is calculated using the formula

\frac{2a}{e}

34 Easy

34 of 392

Analyzing +4 | -1

Let

S = \left\{ {\left( {x,\,y} \right) \in {R^2}:\frac{{{y^2}}}{{1 + r}} - \frac{{{x^2}}}{{1 - r}} = 1} \right\}

, where

r \ne \pm 1

. Then,

S

represents

View answer and explanation

Correct option: C

An ellipse whose eccentricity is

\sqrt {\frac{2}{{r + 1}}}

, when

r > 1

Option C (An ellipse whose eccentricity is

\sqrt {\frac{2}{{r + 1}}}

, when

r > 1

) is correct because when r > 1, the equation represents an ellipse, and its eccentricity is calculated as

\sqrt {\frac{2}{{r + 1}}}

35 Easy

35 of 392

Application +4 | -1

The equation of tangent to the ellipse

{x^2} + 4{y^2} = 5

\left( { - 1,1} \right)

View answer and explanation

Correct option: D

x - 4y + 5 = 0

Option D (

x - 4y + 5 = 0

) is correct because it represents the equation of the tangent to the ellipse at the specified point, obtained by direct substitution into the ellipse equation.

36 Easy

36 of 392

Application +4 | -1

The eccentricity of an ellipse is

\frac{1}{3}

, latus rectum is 4 and centre is

(0,0)

, then the equation of the ellipse is (given major axis is along

x

-axis)

View answer and explanation

Correct option: B

\frac{16 x^2}{81}+\frac{2 y^2}{9}=1

Option A (

\frac{16 x^2}{81}+\frac{2 y^2}{9}=1

) is correct because it satisfies the given eccentricity and latus rectum conditions for the ellipse.

37 Easy

37 of 392

Application +4 | -1

y=m x+c

is the tangent on the ellipse

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

, then the value of

c

View answer and explanation

Correct option: C

\pm \sqrt{9 m^2+4}

Option C (

\pm \sqrt{{9 m^2+4}}

) is correct because, for the ellipse

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

, the condition for tangency of the line y = mx + c is

c = \pm \sqrt{{a^2m^2 + b^2}} = \pm \sqrt{{9m^2 + 4}}

38 Easy

38 of 392

Analyzing +4 | -1

The tangent at a point

P(a \cos \theta, b \sin \theta)

on the ellipse

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

meets the auxiliary circle in two points. The chord joining them subtends a right angle at the centre. Then, the eccentricity of the ellipse is given by

View answer and explanation

Correct option: A

\left(1+\sin ^2 \theta\right)^{-1 / 2}

Option A (

\left(1+\sin ^2 \theta\right)^{-1 / 2}

) is correct because it represents the eccentricity derived from the condition that the chord joining the intersection points subtends a right angle at the center.

39 Easy

39 of 392

Application +4 | -1

If for the ellipse

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

y

-axis is the minor axis and the length of the latusrectum is one half of the length of its minor axis, then its eccentricity is

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the eccentricity calculated from the given relationship between the latus rectum and minor axis lengths.

40 Easy

40 of 392

Application +4 | -1

For the ellipse

25{x^2} + 9{y^2} - 150x - 90y + 225 = 0

, the eccentricity

e

is equal to

View answer and explanation

Correct option: C

\frac{4}{5}

Option C (

\frac{4}{5}

) is correct because the eccentricity is calculated using the formula

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

after rewriting the ellipse equation.

41 Easy

41 of 392

Application +4 | -1

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

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

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents the locus of the midpoints of focal chords of the ellipse, derived by substituting the focus coordinates into the chord equation.

42 Easy

42 of 392

Application +4 | -1

The number of values of c such that the line

y = 4x + c

touches the curve

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

View answer and explanation

Correct option: B

Option B (2) is correct because there are two possible values of 'c' that satisfy the tangency condition for the ellipse.

43 Easy

43 of 392

Understanding +4 | -1

The number of maximum normals which can be drawn from a point to ellipse is

View answer and explanation

Correct option: A

The correct answer is A (4) because a maximum of four normals can be drawn from a point to an ellipse.

44 Easy

44 of 392

Application +4 | -1

The equation of the ellipse with its centre at

(1,2)

, one focus at

(6,2)

and passing through

(4,6)

, is

View answer and explanation

Correct option: A

\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1

Option A (

\frac{(x-1)^2}{45}+\frac{(y-2)^2}{20}=1

) is correct because it satisfies the given conditions: center at (1,2), focus at (6,2), and passing through (4,6).

45 Easy

45 of 392

Understanding +4 | -1

Which one of the following statements is not a false statement?

View answer and explanation

Correct option: B

q

: Circle is a particular case of an ellipse

The correct answer is 'Circle is a particular case of an ellipse' because a circle can be represented as an ellipse with equal major and minor axes.

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46 Easy

46 of 392

Application +4 | -1

P C P^{\prime}

and

D C D^{\prime}

form a pair of conjugate diameters of the ellipse

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

and

R

is any point on the circle

x^2+y^2=c^2

, then

View answer and explanation

Correct option: D

P R^2+D R^2+P^{\prime} R^2+D^{\prime} R^2=2\left(a^2+b^2+2 c^2\right)

Option D (

P R^2 + D R^2 + P' R^2 + D' R^2 = 2(a^2 + b^2 + 2 c^2)

) is correct because it represents the sum of squared distances from points on the ellipse and circle.

47 Easy

47 of 392

Application +4 | -1

The eccentricity of the conic

36{x^2} + 144{y^2} - 36x - 96y - 119 = 0

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it's the eccentricity calculated after rewriting the conic equation in standard form and applying the eccentricity formula.

48 Easy

48 of 392

Application +4 | -1

If the area of the auxiliary circle of the ellipse

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

is twice the area of the ellipse, then the eccentricity of the ellipse is

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it is the calculated eccentricity of the ellipse given the area condition.

49 Easy

49 of 392

Application +4 | -1

Let

e_1

and

e_2

be the eccentricities of the ellipse

\frac{x^2}{18}+\frac{y^2}{4}=1

and hyperbola

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

respectively. If the point

\left(e_1, e_2\right)

lies on the ellipse

15 x^2+3 y^2=k

, then the value of

k

View answer and explanation

Correct option: C

Option C (16) is correct because it is the value obtained by substituting the eccentricities of the ellipse and hyperbola into the given equation.

50 Easy

50 of 392

Application +4 | -1

If the line

x\cos \alpha + y\sin \mathrm{a} = p

be normal to the ellipse

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

, then

View answer and explanation

Correct option: D

{p^2}({a^2}{\sec ^2}\alpha + {b^2}cose{c^2}\alpha ) = {\left( {{a^2} - {b^2}} \right)^2}

Option D (

{p^2}({a^2}{\sec ^2}\alpha + {b^2}cose{c^2}\alpha ) = {\left( {{a^2} - {b^2}} \right)^2}

) is correct because it represents the condition for the given line to be normal to the ellipse.

51 Easy

51 of 392

Application +4 | -1

The equation of the ellipse whose axes are along the coordinate axes, vertices are

(\pm 5,0)

are foci at

(\pm 4,0)

, is

View answer and explanation

Correct option: B

9 x^2+25 y^2=225

Option B (

9 x^2+25 y^2=225

) is correct because it represents the equation of the ellipse with the given vertices and foci, derived by finding the values of a and b.

52 Easy

52 of 392

Application +4 | -1

If tangents drawn to the ellipse through point

(1,2 \sqrt{3})

to the ellipse

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

are at right angle, then value of

b

View answer and explanation

Correct option: C

Option C (2) is correct because the point must lie on the director circle of the ellipse, and substituting the point into the director circle equation allows us to solve for b.

53 Easy

53 of 392

Application +4 | -1

The eccentric angle of a point on the ellipse

9 x^2+4 y^2=36

at a distance

\frac{\sqrt{31}}{2}

units from the centre of the ellipse is '

\theta

', then the value of

|2 \cos \theta|

is-

View answer and explanation

Correct option: A

Option A (1) is correct. By substituting the parametric coordinates of a point on the ellipse and using the given distance from the center, the value of |2cosθ| can be determined.

54 Easy

54 of 392

Application +4 | -1

The distance between the foci of an ellipse is 16 and eccentricity is

1/2

. Length of the major axis of the ellipse is

View answer and explanation

Correct option: D

Option D (32) is correct because the length of the major axis (2a) is calculated using the distance between foci (2ae) and the eccentricity (e).

55 Easy

55 of 392

Application +4 | -1

A set of points is such that each point is three times as far away from the

y

-axis as it is from the point

\left( {4,0} \right)

. Then, the locus of the points is

View answer and explanation

Correct option: C

Ellipse

Option C (Ellipse) is correct because the ratio of the distance from a point to a fixed point (focus) and the distance to a fixed line (directrix) is constant and less than 1, which defines an ellipse.

56 Easy

56 of 392

Application +4 | -1

\frac{x^2}{r^2-r-6}+\frac{y^2}{r^2-6 r+5}=1

will represents the ellipse, if

r

lies in the interval

View answer and explanation

Correct option: C

(5, \infty)

Option C (

(5, \infty)

) is correct because the ellipse equation requires both denominators to be positive, leading to the intersection of the solution sets

r < -2

r > 3

and

r < 1

r > 5

57 Easy

57 of 392

Application +4 | -1

For an ellipse with eccentricity

\frac{1}{2}

the centre is at the origin. If one directrix is

x = 4

, then the equation of the ellipse is

View answer and explanation

Correct option: B

3{x^2} + 4{y^2} = 12

Option B (

3{x^2} + 4{y^2} = 12

) is correct because given the eccentricity and directrix, we can determine the values of a and b, and thus the equation of the ellipse.

58 Easy

58 of 392

Application +4 | -1

The length of the latusrectum of the ellipse

\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{49}} = 1

View answer and explanation

Correct option: B

\frac{{72}}{7}

Option B (

\frac{{72}}{7}

) is correct because the length of the latus rectum of the ellipse is calculated as

\frac{{2{a^2}}}{b} = \frac{{72}}{7}

59 Easy

59 of 392

Application +4 | -1

If any tangent to the ellipse

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

cuts off intercepts of length

h

and

k

on the axes, then find

\frac{a^2}{h^2}+\frac{b^2}{k^2}

View answer and explanation

Correct option: B

Option B (1) is correct because the tangent at any point (a cos θ, b sin θ) to the ellipse cuts off intercepts h and k such that

\frac{{a^2}}{{h^2}} + \frac{{b^2}}{{k^2}} = 1

60 Easy

60 of 392

Application +4 | -1

The equation of the ellipse whose axes are along the coordinate axes, the length of the latus rectum is 5 and the eccentricity is

\frac{2}{3}

View answer and explanation

Correct option: B

\frac{x^2}{81}+\frac{y^2}{45}=\frac{1}{4}

Option B (

\frac{x^2}{81}+ \frac{y^2}{45}=\frac{1}{4}

) is correct because it satisfies the given conditions for the ellipse.

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