Hyperbola — Page 3 | JEE Mathematics

Q21

If e1 and e2 are the eccentricities of the ellipse,

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

and the hyperbola,

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

respectively and (e1, e2) is a point on the ellipse, 15x2 + 3y2 = k, then k is equal to :

A 17

B 16

C 15

D 14

Correct Answer

Option B

Solution

e1 =

\sqrt {1 - {4 \over {18}}}

=

{{\sqrt 7 } \over 3}

e1 =

\sqrt {1 + {4 \over 9}}

=

{{\sqrt {13} } \over 3}

$\because$ (e1, e2 ) lies on 15x2 + 3y2 = k $\therefore$

15\left( {{7 \over 9}} \right) + 3\left( {{{13} \over 9}} \right)

= k $\Rightarrow$ k = 16

Q22

A hyperbola passes through the foci of the ellipse

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

and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is :

A

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

B

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

C

{{{x^2}} \over 9} - {{{y^2}} \over 25} = 1

D x2

-

y2 = 9

Correct Answer

Option B

Solution

{e_1} = \sqrt {1 - {{16} \over {25}}} = {3 \over 5}

foci ( $\pm$ ae, 0) Foci = ( $\pm$ 3, 0) Let equation of hyperbola be

{{{x^2}} \over {{A^2}}} - {{{y^2}} \over {{B^2}}} = 1

Passes through ( $\pm$ 3, 0) A2 = 9, A = 3,

{e_2} = {5 \over 3}

{e_2}^2 = 1 + {{{B^2}} \over {{A^2}}}

{{25} \over 9} = 1 + {{{B^2}} \over 9} \Rightarrow {B^2} = 16

Equation of the hyperbola

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

Q23

The locus of the midpoints of the chord of the circle, x2 + y2 = 25 which is tangent to the hyperbola,

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

is :

A (x2 + y2)2

-

9x2 + 16y2 = 0

B (x2 + y2)2

-

9x2 + 144y2 = 0

C (x2 + y2)2

-

16x2 + 9y2 = 0

D (x2 + y2)2

-

9x2

-

16y2 = 0

Correct Answer

Option A

Solution

tangent of hyperbola

y = mx \pm \sqrt {9{m^2} - 16}

..... (i) which is a chord of circle with mid-point (h, k) so equation of chord T = S1 hx + ky = h2 + k2

y = - {{hx} \over k} + {{{h^2} + {k^2}} \over k}

..... (ii) by (i) and (ii)

m = - {h \over k}

and

\sqrt {9{m^2} - 16}

=

{{{h^2} + {k^2}} \over k}

9{{{h^2}} \over {{k^2}}} - 16 = {{{{\left( {{h^2} + {k^2}} \right)}^2}} \over {{k^2}}}

$\therefore$ Locus : 9x2 $-$ 16y2 = (x2 + y2)2

Q24

Consider a hyperbola H : x2

-

2y2 = 4. Let the tangent at a point P(4,

{\sqrt 6 }

) meet the x-axis at Q and latus rectum at R(x1, y1), x1 > 0. If F is a focus of H which is nearer to the point P, then the area of

\Delta

QFR is equal to :

A

{\sqrt 6 }

-

1

B

{7 \over {\sqrt 6 }}

-

2

C

{4\sqrt 6 }

-

1

D

{4\sqrt 6 }

Correct Answer

Option B

Solution

Given,

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

\Rightarrow {{{x^2}} \over 4} - {{{y^2}} \over 2} = 1 \Rightarrow {{{x^2}} \over {{{(2)}^2}}} - {{{y^2}} \over {{{(\sqrt 2 )}^2}}} = 1

Here, a = 2,

b = \sqrt 2

$\therefore$

e = \sqrt {1 + {{{b^2}} \over {{a^2}}}} = \sqrt {1 + {2 \over 4}} = \sqrt {1 + {1 \over 2}} = \sqrt {{3 \over 2}}

So, Focus (F) = ( $\pm$ a e, 0) = ( $\pm$

\sqrt 6

, 0) Now, equation of tangent at

P(4,\sqrt 6 )

is

x{x_1} - 2y{y_1} = 4

\Rightarrow x\,.\,4 - 2y\,.\,\sqrt 6 = 4

\Rightarrow 4x - 2\sqrt 6 y = 4

\Rightarrow 2x - \sqrt 6 y = 2

....... (i) Putting y = 0 in Eq. (i), we get x-intercept of tangent i.e. x = 1 $\therefore$ Q $\equiv$ (1, 0) Hence, equation of corresponding latus rectum is

x = \sqrt 6

$\therefore$

R \equiv \left( {\sqrt 6 ,{{2(\sqrt 6 - 1)} \over {\sqrt 6 }}} \right)

[putting

x = \sqrt 6

in Eq. (i), we get

y = {{2(\sqrt 6 - 1)} \over {\sqrt 6 }}

] $\therefore$ Area of

\Delta QFR = {1 \over 2} \times (QF) \times (RF)

= {1 \over 2}(\sqrt 6 - 1) \times {{2(\sqrt 6 - 1)} \over {\sqrt 6 }} = {{{{(\sqrt 6 - 1)}^2}} \over {\sqrt 6 }} = \left( {{7 \over {\sqrt 6 }} - 2} \right)

Q25

Let a line L : 2x + y = k, k > 0 be a tangent to the hyperbola x2

-

y2 = 3. If L is also a tangent to the parabola y2 =

\alpha

x, then

\alpha

is equal to :

A 12

B

-

12

C 24

D

-

24

Correct Answer

Option D

Solution

Tangent to hyperbola of Slope m = $-$ 2 (given) y = $-$ 2x $\pm$

\sqrt {3(3)}

\left( {y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} } \right)

$\Rightarrow$ y + 2x = $\pm$ 3 $\Rightarrow$ 2x + y = 3 (k > 0) For parabola y2 = $\alpha$ x

y = mx + {\alpha \over {4m}}

\Rightarrow y = - 2x + {\alpha \over { - 8}}

\Rightarrow {\alpha \over { - 8}} = 3

\Rightarrow \alpha = - 24

Q26

The locus of the centroid of the triangle formed by any point P on the hyperbola

16{x^2} - 9{y^2} + 32x + 36y - 164 = 0

, and its foci is :

A

16{x^2} - 9{y^2} + 32x + 36y - 36 = 0

B

9{x^2} - 16{y^2} + 36x + 32y - 144 = 0

C

16{x^2} - 9{y^2} + 32x + 36y - 144 = 0

D

9{x^2} - 16{y^2} + 36x + 32y - 36 = 0

Correct Answer

Option A

Solution

Given hyperbola is

16{(x + 1)^2} - 9{(y - 2)^2} = 164 + 16 - 36 = 144

\Rightarrow {{{{(x + 1)}^2}} \over 9} - {{{{(y - 2)}^2}} \over {16}} = 1

Eccentricity,

e = \sqrt {1 + {{16} \over 9}} = {5 \over 3}

$\Rightarrow$ foci are (4, 2) and ( $-$ 6, 2) Let the centroid be (h, k) & A( $\alpha$ , $\beta$ ) be point on hyperbola.

So,

h = {{\alpha - 6 + 4} \over 3},k = {{\beta + 2 + 2} \over 3}

\Rightarrow \alpha = 3h + 2,\beta = 3k - 4

( $\alpha$ , $\beta$ ) lies on hyperbola so

16{(3h + 2 + 1)^2} - 9{(3k - 4 - 2)^2} = 144

\Rightarrow 144{(h + 1)^2} - 81{(k - 2)^2} = 144

\Rightarrow 16({h^2} + 2h + 1) - 9({k^2} - 4k + 4) = 16

\Rightarrow 16{x^2} - 9{y^2} + 32x + 36y - 36 = 0

Q27

The point

P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)

lies on the hyperbola

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

having eccentricity

{{\sqrt 5 } \over 2}

. If the tangent and normal at P to the hyperbola intersect its conjugate axis at the point Q and R respectively, then QR is equal to :

A

4\sqrt 3

B 6

C

6\sqrt 3

D

3\sqrt 6

Correct Answer

Option C

Solution

P\left( { - 2\sqrt 6 ,\sqrt 3 } \right)

lies on hyperbola

\Rightarrow {{24} \over {{a^2}}} - {3 \over {{b^2}}} = 1

...... (i)

e = {{\sqrt 5 } \over 2} \Rightarrow {b^2} = {a^2}\left( {{5 \over 4} - 1} \right) \Rightarrow 4{b^2} = {a^2}

Put in (i)

\Rightarrow {6 \over {{b^2}}} - {3 \over {{b^2}}} = 1 \Rightarrow b = \sqrt 3

\Rightarrow a = \sqrt {12}

Tangent at P :

{{ - x} \over {\sqrt 6 }} - {y \over {\sqrt 3 }} = 1 \Rightarrow Q(0,\sqrt 3 )

Slope of

T = - {1 \over {\sqrt 2 }}

Normal at P :

y - \sqrt 3 = \sqrt 2 (x + 2\sqrt 6 )

\Rightarrow R = (0,5\sqrt 3 )

QR = 6\sqrt 3

Q28

The locus of the mid points of the chords of the hyperbola x2

-

y2 = 4, which touch the parabola y2 = 8x, is :

A y3(x

-

2) = x2

B x3(x

-

2) = y2

C y2(x

-

2) = x3

D x2(x

-

2) = y3

Correct Answer

Option C

Solution

T = S1 xh $-$ yk = h2 $-$ k2

y = {{xh} \over k} - {{({h^2} - {k^2})} \over k}

this touches y2 = 8x then

c = {a \over m}

\left( {{{{k^2} - {h^2}} \over k}} \right) = {{2k} \over h}

2y2 = x(y2 $-$ x2) y2(x $-$ 2) = x3

Q29

Let a > 0, b > 0. Let e and l respectively be the eccentricity and length of the latus rectum of the hyperbola

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

. Let e' and l' respectively be the eccentricity and length of the latus rectum of its conjugate hyperbola. If

{e^2} = {{11} \over {14}}l

and

{\left( {e'} \right)^2} = {{11} \over 8}l'

, then the value of

77a + 44b

is equal to :

A 100

B 110

C 120

D 130

Correct Answer

Option D

Solution

H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

, then

{e^2} = {{11} \over {14}}l

(l be the length of LR)

\Rightarrow {a^2} + {b^2} = {{11} \over 7}{b^2}a

..... (i) and

e{'^2} = {{11} \over 8}l'

(l' be the length of LR of conjugate hyperbola)

\Rightarrow {a^2} + {b^2} = {{11} \over 4}{a^2}b

....... (ii) By (i) and (ii)

7a = 4b

then by (i)

{{16} \over {49}}{b^2} + {b^2} = {{11} \over 7}{b^2}\,.\,{{4b} \over 7}

\Rightarrow 44b = 65

and

77a = 65

$\therefore$

77a + 44b = 130

Q30

Let the eccentricity of the hyperbola

H:{{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1

be

\sqrt {{5 \over 2}}

and length of its latus rectum be

6\sqrt 2

. If

y = 2x + c

is a tangent to the hyperbola H, then the value of c2 is equal to :

A 18

B 20

C 24

D 32

Correct Answer

Option B

Solution

1 + {{{b^2}} \over {{a^2}}} = {5 \over 2} \Rightarrow {{{b^2}} \over {{a^2}}} = {3 \over 2}

{{2{b^2}} \over a} = 6\sqrt 2 \Rightarrow 2.\,{3 \over 2}.\,a = 6\sqrt 2

\Rightarrow a = 2\sqrt 2 ,\,{b^2} = 12

{c^2} = {a^2}{m^2} - {b^2} = 8.4 - 12 = 20