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JEE • MATHEMATICS • HYPERBOLA

Hyperbola 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 13

Application +4 | -1

If eccentricity of conjugate hyperbola of the hyperbola

\left|\sqrt{(x-1)^2+(y-2)^2}-\sqrt{(x-5)^2+(y-5)^2}\right|

=3

e

, then the value of

8 \mathrm{e}

is equal to

View answer and explanation

Correct option: A

Option A (10) is correct because the eccentricity of the conjugate hyperbola is calculated, and then 8e is found to be 10.

2 Hard

2 of 13

Analyzing +4 | -1

The circle

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

and hyperbola

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

intersect at the points

A

and

B

. The equation of the circle with

A B

as its diameter, is

View answer and explanation

Correct option: A

x^2+y^2-12 x+24=0

Option A (

x^2+y^2-12 x+24=0

) is correct because it represents the equation of the circle with AB as its diameter, where A and B are the intersection points of the given circle and hyperbola.

3 Hard

3 of 13

Analyzing +4 | -1

If the chords of the rectangular hyperbola

{x^2} - {y^2} = {a^2}

touch the parabola

{y^2} = 4ax,

then the locus of their mid-points is

View answer and explanation

Correct option: B

{y^2}\left( {x - a} \right) = {x^3}

Option B (

{y^2}\left( {x - a} \right) = {x^3}

) is correct because it represents the locus of the mid-points of the chords of the rectangular hyperbola that touch the given parabola.

4 Hard

4 of 13

Application +4 | -1

If the tangent at

\left( {\alpha ,\beta } \right)

to the hyperbola

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

cuts the auxiliary circle at points whose ordinates are

{y_1}

and

{y_2}

, then

\frac{1}{{{y_1}}} + \frac{1}{{{y_2}}} =

View answer and explanation

Correct option: D

\frac{2}{\beta }

Option D (

\frac{2}{\beta }

) is correct because it is derived from finding the intersection of the tangent to the hyperbola and its auxiliary circle, then using Vieta's formulas.

5 Hard

5 of 13

Analyzing +4 | -1

The normal at

P

to a hyperbola of eccentricity

e

, intersects its transverse and conjugate axes at

L

and

M

respectively. If locus of the mid point of

LM

is hyperbola, then eccentricity of the hyperbola is

View answer and explanation

Correct option: B

\frac{e}{{\sqrt {\left( {{e^2} - 1} \right)} }}

Option B (

\frac{e}{{\sqrt {\left( {{{e^2} - 1} \right)} }}

) is correct because it is derived from the geometry of the hyperbola and the properties of the normal line intersecting the transverse and conjugate axes.

6 Hard

6 of 13

Analyzing +4 | -1

The normal at a point

P

on the ellipse

{x^2} + 4{y^2} = 16

meets the

x - axis

Q

. If

M

is the mid point of the line segment

PQ

, then the locus of

M

intersects the latusrectum of the given ellipse at the points

View answer and explanation

Correct option: C

\left( { \pm 2\sqrt 3 , \pm \frac{1}{7}} \right)

Option D (

\left( { \pm 2\sqrt 3 , \pm \frac{1}{7}} \right)

) is correct because the locus of M intersects the latus rectum of the ellipse at these points.

7 Hard

7 of 13

Application +4 | -1

The product of the perpendicular from two foci on any tangent to the hyperbola

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

, is

View answer and explanation

Correct option: B

{b^2}

Option B (b²) is correct because the product of the perpendiculars from the foci to any tangent of the hyperbola is equal to the square of the semi-minor axis.

8 Hard

8 of 13

Application +4 | -1

Tangents at any points on the hyperbola

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

cut the axes at

A

and

B

respectively. If the rectangle

OAPB,

where

O

is the origin is completed, then locus of point

P

is given by

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because it represents the locus of point P, where tangents to the hyperbola cut the axes at A and B, and OAPB is a rectangle.

9 Hard

9 of 13

Analyzing +4 | -1

Let

F_1, F_2

are the foci of the hyperbola

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

and

F_3, F_4

are the foci of its conjugate hyperbola. If

\mathrm{e}_{\mathrm{H}}

and

\mathrm{e}_{\mathrm{C}}

are eccentricities of hyperbola and its conjugate hyperbola respectively, then the statement which holds TRUE is :

View answer and explanation

Correct option: C

Area of the quadrilateral formed by their foci is 50 sq. units.

Option C (Area of the quadrilateral formed by their foci is 50 sq. units) is correct because the area of the quadrilateral formed by the foci of the hyperbola and its conjugate is indeed 50 square units.

10 Hard

10 of 13

Analyzing +4 | -1

Let

\mathrm{P}(3 \sec \theta, 2 \tan \theta)

and

\mathrm{Q}(3 \sec \phi, 2 \tan \phi)

where

\theta+\phi=\frac{\pi}{2}

, be two distinct points on the hyperbola

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

. Then the ordinate of the point of intersection of the normals at

\mathrm{P}

and

\mathrm{Q}

is:

View answer and explanation

Correct option: D

-\frac{13}{2}

Option D (

-\frac{13}{2}

) is correct because it is the ordinate of the intersection point of the normals at P and Q, derived using the properties of hyperbolas and trigonometric identities.

11 Hard

11 of 13

Application +4 | -1

If the angle between the asymptotes of the hyperbola

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

2 \theta

and

e

is its eccentricity then

\sin \theta

is equal to

View answer and explanation

Correct option: D

\sqrt{1-\frac{1}{e^2}}

Option D (

\sqrt{1-\frac{1}{e^2}}

) is correct because it correctly relates the angle between asymptotes and the eccentricity of the hyperbola.

12 Hard

12 of 13

Analyzing +4 | -1

Let

P(a\sec \mathrm{\theta },b\tan \mathrm{\theta })

and

Q(a\sec \phi ,b\tan \phi ),

where

\mathrm{\theta } + \phi = \frac{\mathrm{\pi }}{2}

be two points on the hyperbola

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

. If

\left( {h,k} \right)

is the point of intersection of normals at

P

and

Q

, then

k

is equal to

View answer and explanation

Correct option: D

- \left[ {\frac{{{a^2} + {b^2}}}{b}} \right]

Option D (

\ - \left[ {\frac{{{{a^2} + {b^2}}}}{b}} \right]

) is correct because it is the y-coordinate of the intersection point of the normals at P and Q, derived using the given condition θ + φ = π/2.

13 Hard

13 of 13

Understanding +4 | -1

If a diameter of a hyperbola meets the hyperbola in real points then

View answer and explanation

Correct option: A

It meets the conjugate hyperbola in imaginary points.

The correct answer is A (It meets the conjugate hyperbola in imaginary points.) because if a diameter intersects a hyperbola in real points, it intersects its conjugate in imaginary points.

Revision notes and key takeaways

Review HYPERBOLA in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

✓ This set contains 13 single-correct-answer MCQs on HYPERBOLA.
✓ 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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