Hyperbola — Page 2 | JEE Mathematics

Q11

If 5x + 9 = 0 is the directrix of the hyperbola 16x2 – 9y2 = 144, then its corresponding focus is :

A

\left( {{5 \over 3},0} \right)

B (5, 0)

C (- 5, 0)

D

\left( { - {5 \over 3},0} \right)

Correct Answer

Option C

Solution

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

$\therefore$ a = 3 and b = 4

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

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

$\Rightarrow$ e =

5 \over 3

$\therefore$ focus is (–ae, 0) = (–5, 0)

Q12

If the line y = mx + 7

\sqrt 3

is normal to the hyperbola

{{{x^2}} \over {24}} - {{{y^2}} \over {18}} = 1

, then a value of m is :

A

{3 \over {\sqrt 5 }}

B

{{\sqrt 15 } \over 2}

C

{{\sqrt 5 } \over 2}

D

{2 \over {\sqrt 5 }}

Correct Answer

Option D

Solution

Given line y = mx + 7

\sqrt 3

.....(1) Given hyperbola

{{{x^2}} \over {24}} - {{{y^2}} \over {18}} = 1

Here

{a^2} = 24

and

{b^2} = 18

We know the equation of normal to the hyperbola is

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

$\Rightarrow$

y = mx \pm {{m\left( {42} \right)} \over {\sqrt {24 - 18{m^2}} }}

.....(2) Comparing (1) and (2), we get

{{m\left( {42} \right)} \over {\sqrt {24 - 18{m^2}} }}

=

7\sqrt 3

$\Rightarrow$ 36m2 = 72 - 54m2 $\Rightarrow$ 90m2 = 72 $\Rightarrow$ m2 =

{{72} \over {90}}

$\Rightarrow$ m =

{2 \over {\sqrt 5 }}

Q13

If the eccentricity of the standard hyperbola passing through the point (4,6) is 2, then the equation of the tangent to the hyperbola at (4,6) is :

A 2x – y – 2 = 0

B 3x – 2y = 0

C 2x – 3y + 10 = 0

D x – 2y + 8 = 0

Correct Answer

Option A

Solution

Formula for standard hyperbola :

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

It passes through (4, 6) $\therefore$

{{16} \over {{a^2}}} - {{36} \over {{b^2}}} = 1

........(1) We know,

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

$\Rightarrow$ 4 =

1 + {{{b^2}} \over {{a^2}}}

[ as e = 2 ] $\Rightarrow$

{{b^2} = 3{a^2}}

....(2) Putting

{{b^2} = 3{a^2}}

in equation (1)

{{16} \over {{a^2}}} - {{36} \over {3{a^2}}} = 1

$\Rightarrow$

{{a^2} = 4}

$\therefore$ ,

{{b^2} = 12}

So Equation of hyperbola is

{{{x^2}} \over 4} - {{{y^2}} \over {12}} = 1

Equation of tangent to the hyperbola at (4, 6) is

{{4x} \over 4} - {{6y} \over {12}} = 1

$\Rightarrow$

x - {y \over 2} = 1

$\Rightarrow$ 2x – y – 2 = 0

Q14

If the vertices of a hyperbola be at (–2, 0) and (2, 0) and one of its foci be at (–3, 0), then which one of the following points does not lie on this hyperbola?

A

\left( {6,5\sqrt 2 } \right)

B

\left( {2\sqrt 6 ,5} \right)

C

\left( { - 6,2\sqrt {10} } \right)

D

\left( {4,\sqrt {15} } \right)

Correct Answer

Option A

Solution

ae = 3, e =

{3 \over 2}

, b2 = 4

\left( {{9 \over 4} - 1} \right)

, b2 = 5

{{{x^2}} \over 4} - {{{y^2}} \over 5} = 1

Q15

If a hyperbola has length of its conjugate axis equal to 5 and the distance between its foci is 13, then the eccentricity of the hyperbola is :

A

{{13} \over 6}

B 2

C

{{13} \over 12}

D

{{13} \over 8}

Correct Answer

Option C

Solution

2b = 5 and 2ae = 13 b2 = a2(e2 $-$ 1) $\Rightarrow$

{{25} \over 4}

=

{{169} \over 4}

$-$ a2 $\Rightarrow$ a $=$ 6 $\Rightarrow$ e $=$

{{13} \over {12}}

Q16

Let P(3, 3) be a point on the hyperbola,

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

. If the normal to it at P intersects the x-axis at (9, 0) and e is its eccentricity, then the ordered pair (a2, e2) is equal to :

A

\left( {{9 \over 2},2} \right)

B

\left( {{3 \over 2},2} \right)

C (9,3)

D

\left( {{9 \over 2},3} \right)

Correct Answer

Option D

Solution

Given hyperbola,

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

Point P (3, 3) is on the parabola $\therefore$

{9 \over {{a^2}}} - {9 \over {{b^2}}} = 1

...(1) Equation of normal at (x1, y1),

{{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2}{e^2}

Normal at p(3, 3),

{{{a^2}x} \over 3} - {{{b^2}y} \over 3} = {a^2}{e^2}

... (2) It intersect x-axis at (9, 0), Putting in equation (2),

{{9{a^2}} \over 3} - 0 = {a^2}{e^2}

\Rightarrow 3{a^2} = {a^2}{e^2}

\Rightarrow {e^2} = 3

Also,

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

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

\Rightarrow {b^2} = 2{a^2}

Putting the value of b2 in equation (1),

{9 \over {{a^2}}} - {9 \over {2{a^2}}} = 1

\Rightarrow {a^2} = {9 \over 2}

$\therefore$

({a^2},{e^2}) = \left( {{9 \over 2},3} \right)

Q17

If the line y = mx + c is a common tangent to the hyperbola

{{{x^2}} \over {100}} - {{{y^2}} \over {64}} = 1

and the circle x2 + y2 = 36, then which one of the following is true?

A 5m = 4

B 8m + 5 = 0

C c2 = 369

D 4c2 = 369

Correct Answer

Option D

Solution

{{{x^2}} \over {100}} - {{{y^2}} \over {64}} = 1

$\therefore$ c = $\pm$

\sqrt {{a^2}{m^2} - {b^2}}

$\Rightarrow$ c = $\pm$

\sqrt {100{m^2} - 64}

General tangent to hyperbola in slope form is y = mx $\pm$

\sqrt {100{m^2} - 64}

This tangent is also tangent to the circle x2 + y2 = 36, whose center (0, 0) and radius = 6.

Distance of the tangent from the center is

\left| {{{\sqrt {100{m^2} - 64} } \over {\sqrt {{m^2} + 1} }}} \right|

= 6 $\Rightarrow$ 100m2 — 64 = 36m2 + 36 $\Rightarrow$ 64m2 = 100 $\Rightarrow$ m =

{{10} \over 8}

$\therefore$ c2 = 100 $\times$

{{100} \over {64}}

- 64 $\Rightarrow$ c2 =

{{{{100}^2} - {{64}^2}} \over {64}}

$\Rightarrow$ c2 =

{{164 \times 36} \over {64}}

$\Rightarrow$ 4c2 = 369

Q18

Let e1 and e2 be the eccentricities of the ellipse,

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

(b < 5) and the hyperbola,

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

respectively satisfying e1e2 = 1. If

\alpha

and

\beta

are the distances between the foci of the ellipse and the foci of the hyperbola respectively, then the ordered pair (

\alpha

,

\beta

) is equal to :

A (8, 10)

B (8, 12)

C

\left( {{{24} \over 5},10} \right)

D

\left( {{{20} \over 3},12} \right)

Correct Answer

Option A

Solution

For ellipse

{{{x^2}} \over {25}} + {{{y^2}} \over {{b^2}}} = 1\,\,\,(b < 5)

Let e1 is eccentricity of ellipse $\therefore$ b2 = 25 (1 $-$

{e_1}^2

) ........(1) Again for hyperbola

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

Let e2 is eccentricity of hyperbola. $\therefore$

{b^2} = 16({e_1}^2 - 1)

......(2) by (1) & (2)

25(1 - {e_1}^2)\, = \,16({e_1}^2 - 1)

Now e1 . e2 = 1 (given) $\therefore$

25(1 - {e_1}^2)\, = \,16\left( {{{1 - {e_1}^2} \over {{e_1}^2}}} \right)

or e1 =

{4 \over 5}

$\therefore$ e2 =

{5 \over 4}

Now distance between foci is 2ae $\therefore$ Distance for ellipse =

2 \times 5 \times {4 \over 5} = 8 = \alpha

Distance for hyperbola =

2 \times 4 \times {5 \over 4} = 10 = \beta

$\therefore$ ( $\alpha$ , $\beta$ ) $\equiv$ (8, 10)

Q19

A hyperbola having the transverse axis of length

\sqrt 2

has the same foci as that of the ellipse 3x2 + 4y2 = 12, then this hyperbola does not pass through which of the following points?

A

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

B

\left( {\sqrt {{3 \over 2}} ,{1 \over {\sqrt 2 }}} \right)

C

\left( { - \sqrt {{3 \over 2}} ,1} \right)

D

\left( {{1 \over {\sqrt 2 }},0} \right)

Correct Answer

Option B

Solution

Ellipse :

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

eccentricity =

\sqrt {1 - {3 \over 4}} = {1 \over 2}

$\therefore$ foci = ( $\pm$ 1, 0) for hyperbola, given

2a = \sqrt 2 \Rightarrow a = {1 \over {\sqrt 2 }}

$\therefore$ hyperbola will be

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

eccentricity =

\sqrt {1 + 2{b^2}}

$\therefore$ foci

= \left( { \pm \sqrt {{{1 + 2{b^2}} \over 2}} ,0} \right)

$\because$ Ellipse and hyperbola have same foci

\Rightarrow \sqrt {{{1 + 2{b^2}} \over 2}} = 1

\Rightarrow {b^2} = {1 \over 2}

$\therefore$ Equation of hyperbola :

{{{x^2}} \over {1/2}} - {{{y^2}} \over {1/2}} = 1

$\Rightarrow$

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

Clearly

\left( {\sqrt {{3 \over 2}} ,{1 \over {\sqrt 2 }}} \right)

does not lie on it.

Q20

A line parallel to the straight line 2x – y = 0 is tangent to the hyperbola

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

at the point

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

. Then

x_1^2 + 5y_1^2

is equal to :

A 5

B 6

C 10

D 8

Correct Answer

Option B

Solution

Tangent of hyperbola

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

at the point (x1, y1) is

{{x{x_1}} \over 4} - {{y{y_1}} \over 2} = 1

which is parallel to 2x – y = 0 $\therefore$ Slope of tangent

{{x{x_1}} \over 4} - {{y{y_1}} \over 2} = 1

= Slope of 2x – y = 0 $\Rightarrow$

{{{x_1}} \over {2{y_1}}}

= 2 $\Rightarrow$ x1 = 4y1 ....(1) (x1, y1) lies on hyperbola $\therefore$

{{x_1^2} \over 4} - {{y_1^2} \over 2} = 1

....(2) From (1) & (2)

{{{{\left( {4{y_1}} \right)}^2}} \over 4} - {{y_1^2} \over 2} = 1

$\Rightarrow$

4y_1^2 - {{y_1^2} \over 2} = 1

$\Rightarrow$

7y_1^2 = 2

$\Rightarrow$

y_1^2 = {2 \over 7}

Now

x_1^2 + 5y_1^2

=

{\left( {4{y_1}} \right)^2} + {\left( {5{y_1}} \right)^2}

= 21

y_1^2

= 21

\times {2 \over 7}

= 6