If the eccentricity of the hyperbola $x^2-y^2 \operatorname{coces}^2 \alpha=25$ is $\sqrt{5}$ times the eccentricity of the ellipse $x^2 \operatorname{cosec}^2 \alpha+y^2=5$, then $\alpha$ is equal to :

$\sin ^{-1} \sqrt{\frac{3}{4}}$

$\tan ^{-1} \sqrt{\frac{2}{5}}$

$\sin ^{-1} \sqrt{\frac{2}{5}}$

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JEE • MATHEMATICS • ELLIPSE

Ellipse practice · Mixed difficulty

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571 Medium

571 of 594

Application +4 | -1

The centre of the ellipse

9{x^2} + 25{y^2} - 18x - 100y - 166 = 0,

View answer and explanation

Correct option: D

\left( {1,2} \right)

Option D ((1,2)) is correct because completing the square allows us to rewrite the ellipse equation and identify the center.

572 Easy

572 of 594

Application +4 | -1

Let

x = 4

be a directrix to an ellipse whose centre is at origin and its eccentricity is

\frac{1}{2}\,.

P\left( {1,\,\beta } \right),\,\,\,\beta > 0\,

is a point on this ellipse, then the equation of the normal to it at

P

View answer and explanation

Correct option: D

4x - 2y = 1

The correct answer is D (

4x - 2y = 1

) because it is the equation of the normal to the ellipse at point P, derived using the ellipse's properties and the given directrix and eccentricity.

573 Easy

573 of 594

Analyzing +4 | -1

Given

f

is increasing, the equation

\frac{x^2}{f(2 a)}+\frac{y^2}{f\left(a^2-3\right)}=1

represents an ellipse with

X

-axis as major axis if

View answer and explanation

Correct option: C

(-1,3)

Option C

(-1,3)

is correct because it represents the interval where

a

satisfies the inequality derived from the condition for the ellipse to have the X-axis as the major axis.

574 Easy

574 of 594

Application +4 | -1

A point on the ellipse

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

at a distance equal to the mean of lengths of the semi-major and semi-minor axis from the centre, is

View answer and explanation

Correct option: A

\left(\frac{2 \sqrt{91}}{7}, \frac{3 \sqrt{105}}{14}\right)

Option A (

\left(\frac{2 \sqrt{91}}{7}, \frac{3 \sqrt{105}}{14}\right)

) is correct because substituting the parametric coordinates into the distance equation and solving yields this point.

575 Easy

575 of 594

Remembering +4 | -1

Length of latusrectum of the ellipse

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

, is

View answer and explanation

Correct option: A

\frac{2 a^2}{b}

, if

a < b

Option C (

\frac{2 b^2}{a}

, if

a < b

) is incorrect. Option A (

\frac{2 a^2}{b}

, if

a < b

) is correct because the length of the latus rectum of an ellipse when a < b is

\frac{2a^2}{b}

576 Easy

576 of 594

Understanding +4 | -1

The curve with parametric equations

x = \mathrm{\alpha } + 5\cos \mathrm{\theta },\;y = \mathrm{\beta } + 4\sin \mathrm{\theta }

(where

\theta

is parameter) is

View answer and explanation

Correct option: B

An ellipse

Option B (An ellipse) is correct because the parametric equations describe an ellipse centered at (α, β) with semi-major and semi-minor axes of 5 and 4, respectively.

577 Easy

577 of 594

Application +4 | -1

The eccentricity of the hyperbola whose length of the latusrectum is equal to 8 and the length of its conjugate axis is equal to half of the distance between its foci, is

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is the eccentricity calculated from the given conditions on the latus rectum and conjugate axis.

578 Easy

578 of 594

Application +4 | -1

The eccentric angle of a point on the ellipse

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

, whose distance from the centre of the ellipse is 2, is

View answer and explanation

Correct option: A

\frac{\pi }{4}

The correct answer is A (

\frac{\pi}{4}

) because substituting the parametric form of the ellipse into the distance formula and solving for the eccentric angle yields this result.

579 Easy

579 of 594

Application +4 | -1

If the eccentricity of the hyperbola

x^2-y^2 \operatorname{coces}^2 \alpha=25

\sqrt{5}

times the eccentricity of the ellipse

x^2 \operatorname{cosec}^2 \alpha+y^2=5

, then

\alpha

is equal to :

View answer and explanation

Correct option: A

\tan ^{-1} \sqrt{2}

Option A (

\tan^{-1} \sqrt{2}

) is correct because by equating the eccentricities of the hyperbola and the ellipse as given in the question, we can solve for

\alpha

and find that

\alpha = \tan^{-1} \sqrt{2}

580 Easy

580 of 594

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 minor axis is

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the given condition (latus rectum is half of its minor axis) leads to the relationship

a = 2b

, which allows us to calculate the eccentricity as

\sqrt{{1 - \frac{{b^2}}{{a^2}}}} = \sqrt{{1 - \frac{{1}}{{4}}}} = \frac{{\sqrt{{3}}}}{{2}}

581 Medium

581 of 594

Analyzing +4 | -1

Let

S_1(\alpha, 0)

and

S_2(-\alpha, 0)

be the focii of the ellipse

\frac{x^2}{4}+\frac{y^2}{3}=1,(\alpha > 0)

. If

S=0

be the locus of the image of the focus

S_1

about any tangent on the ellipse, then area bounded by

S=0

and

x+3 y+1=0

is equal to

View answer and explanation

Correct option: C

8 \pi

sq. units

The correct answer is

8 \pi

sq. units because the locus of the image of the focus about any tangent is a circle, and the area bounded by this circle and the given line can be calculated.

582 Easy

582 of 594

Application +4 | -1

For an ellipse

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

with vertices

A

and

A^{\prime}

, tangent drawn at the point

P

in the first quadrant meets the

y

-axis in

\mathrm{Q}

and the chord

\mathrm{A}^{\prime} \mathrm{P}

meets the

\mathrm{y}

-axis in

\mathrm{M}

. If '

\mathrm{O}

' is the origin then

\mathrm{OQ}^2-\mathrm{MQ}^2

equals to

View answer and explanation

Correct option: C

Option B (by = ax - √(a⁴⁺a² b²⁺b⁴)) is correct because it satisfies the condition for a common tangent to both ellipses.

583 Easy

583 of 594

Application +4 | -1

The eccentricity of the ellipse

9{x^2} + 5{y^2} - 18x - 20y - 16 = 0

View answer and explanation

Correct option: B

\frac{2}{3}

Option B (

\frac{2}{3}

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

584 Medium

584 of 594

Analyzing +4 | -1

From point

P(8,27)

tangents

P Q

and

P R

are drawn to the ellipse

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

. Then, angle subtended by

Q R

at origin is

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the calculated angle subtended by QR at the origin does not match any of the given options. The angle is found by homogenizing the ellipse equation.

585 Easy

585 of 594

Application +4 | -1

If the point

p

on the ellipse

4 x^2+5 y^2=20

is farthest from the point

Q(0,-4)

, then

P Q^2

is equal to

View answer and explanation

Correct option: D

Option D (36) is correct because it represents the maximum squared distance between a point on the ellipse and the given point Q(0, -4).

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586 Easy

586 of 594

Application +4 | -1

The length of the chord of the ellipse

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

, where mid-point is

\left(\frac{1}{2}, \frac{2}{3}\right)

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the calculated length of the chord does not match any of the provided options.

587 Medium

587 of 594

Application +4 | -1

The locus of chords of contact of perpendicular tangents to the ellipse

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

touch another fixed ellipse is

View answer and explanation

Correct option: C

\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{\left(a^2+b^2\right)}

Option C (

\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{\left(a^2+b^2\right)}

) is correct because it represents the equation of the fixed ellipse touched by the chords of contact of perpendicular tangents to the given ellipse.

588 Easy

588 of 594

Application +4 | -1

The eccentricity of an ellipse, with centre at the origin is

2/3.

If one of directrices is

x = 6,

then equation of the ellipse is

View answer and explanation

Correct option: C

5x^2 + 9y^2 = 80

Option C (

5x^2 + 9y^2 = 80

) is correct because it is derived by using the given eccentricity and directrix to find the equation of the ellipse.

589 Easy

589 of 594

Analyzing +4 | -1

The curve represented by the equation

4{x^2} + 16{y^2} - 24x - 32y - 12 = 0

View answer and explanation

Correct option: D

An ellipse with eccentricity

\sqrt 3 /2

Option D (An ellipse with eccentricity

\sqrt 3 /2

) is correct because the equation represents an ellipse, and its eccentricity is calculated as

\sqrt{1 - (b^2/a^2)}

590 Easy

590 of 594

Application +4 | -1

If the eccentricities of the two ellipse

\frac{x^2}{169}+\frac{y^2}{25}=1

and

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

are equal, then the value

\frac{a}{b}

, is

View answer and explanation

Correct option: C

\frac{13}{5}

Option C (

\frac{13}{5}

) is correct because it is derived from equating the eccentricities of the two ellipses and solving for the ratio

\frac{a}{b}

591 Medium

591 of 594

Application +4 | -1

The line

3x + 5y = 15\sqrt 2

is a tangent to the ellipse

\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1,

at a point whose eccentric angle is

View answer and explanation

Correct option: B

\frac{\pi }{4}

Option B (

\frac{\pi}{4}

) is correct because by comparing the given tangent equation with the parametric form of the tangent to the ellipse, we can find the eccentric angle.

592 Easy

592 of 594

Application +4 | -1

The equation of the normal to the ellipse

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

at the point

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

View answer and explanation

Correct option: C

\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2

Option C (

\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2

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

593 Medium

593 of 594

Application +4 | -1

Equation of the ellipse with eccentricity

\frac{1}{2}

and foci at

\left( { \pm \,1,0} \right)

View answer and explanation

Correct option: B

\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1

Option B (

\frac{{{x^2}}}{4} + \frac{{{y^2}}}{3} = 1

) is correct because it represents the equation of an ellipse with the given eccentricity and foci.

594 Medium

594 of 594

Application +4 | -1

The radius of the circle passing through the foci of the ellipse

\frac{{{x^2}}}{4} + \frac{4}{7}{y^2} = 1

and having its centre at

\left( {\frac{1}{2},\;2} \right)

, is

View answer and explanation

Correct option: A

\sqrt 5

Option A (

\sqrt 5

) is correct because the radius is calculated using the distance formula between the focus and the center of the circle.

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