Hyperbola — Page 6 | JEE Mathematics

Q51

A hyperbola whose transverse axis is along the major axis of the conic,

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

and has vertices at the foci of this conic. If the eccentricity of the hyperbola is

{3 \over 2},

then which of the following points does NOT lie on it?

A (0, 2)

B

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

C

\left( {\sqrt {10} ,2\sqrt 3 } \right)

D

\left( {5,2\sqrt 3 } \right)

Correct Answer

Option D

Solution

{{{x^2}} \over {12}} + {{{y^2}} \over {16}}

= 1 e =

\sqrt {1 - {{12} \over {16}}}

=

{1 \over 2}

Foci (0, 2) & (0, $-$ 2) So, transverse axis of hyperbola = 2b = 4 $\Rightarrow$ b = 2 & a2 = 12 (e2 $-$ 1) $\Rightarrow$ a2 = 4

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

$\Rightarrow$ a2 = 5 $\therefore$ It's equation is

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

= $-$ 1 The point (5, 2

\sqrt 3

) does not satisfy the above equation.

Q52

The locus of the point of intersection of the straight lines, tx

-

2y

-

3t = 0 x

-

2ty + 3 = 0 (t

\in

R), is :

A an ellipse with eccentricity

{2 \over {\sqrt 5 }}

B an ellipse with the length of major axis 6

C a hyperbola with eccentricity

\sqrt 5

D a hyperbola with the length of conjugate axis 3

Correct Answer

Option D

Solution

Here, tx $-$ 2y $-$ 3t = 0 & x $-$ 2ty + 3 = 0 On solving, we get; y =

{{6t} \over {2{t^2} - 2}}

=

{{3t} \over {{t^2} - 1}}

& x =

{{3{t^2} + 3} \over {{t^2} - 1}}

Put t = tan $\theta$ $\therefore$ x = $-$ 3 sec 2 $\theta$ & 2y = 3 ( $-$ tan 2 $\theta$ ) $\because$ sec22 $\theta$ $-$ tan22 $\theta$ = 1 $\Rightarrow$

{{{x^2}} \over 9}

$-$

{{{y^2}} \over {9/4}}

= 1 which represents at hyperbola $\therefore$ a2 = 9 & b2 = 9/4 $\lambda$ (T.A.) = 6; e2 = 1 +

{{9/4} \over 9}

= 1 +

{1 \over 4}

$\Rightarrow$ e =

{{\sqrt 5 } \over 2}

Q53

If the tangents drawn to the hyperbola 4y2 = x2 + 1 intersect the co-ordinate axes at the distinct points A and B then the locus of the mid point of AB is :

A x2

-

4y2 + 16x2y2 = 0

B x2

-

4y2

-

16x2y2 = 0

C 4x2

-

y2 + 16x2y2 = 0

D 4x2

-

y2

-

16x2y2 = 0

Correct Answer

Option B

Solution

Equation of hyperbola is : 4y2 = x2 + 1 $\Rightarrow$ $-$ x2 + 4y2 = 1 $\Rightarrow$

\,\,\,

$-$

{{{x^2}} \over {{1^2}}}

+

{{{y^2}} \over {{{\left( {{1 \over 2}} \right)}^2}}}

= 1 $\therefore$ a = 1, b =

{1 \over 2}

Now, tangent to the curve at point (x1, y1) is given by 4 $\times$ 2y1

{{dy} \over {dx}}

= 2x1 $\Rightarrow$

{{dy} \over {dx}}

=

{{2{x_1}} \over {8{y_1}}}

=

{{{x_1}} \over {4{y_1}}}

Equation of tangent at (x1, y1) is y = mx + c $\Rightarrow$ y =

{{{x_1}} \over {4{y_1}}}

. x + c As tangent passes through (x1, y1) $\therefore$ y1 =

{{{x_1}{x_1}} \over {4{y_1}}} + c

$\Rightarrow$ C =

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

=

{1 \over {4{y_1}}}

Therefore, y =

{{{x_1}} \over {4{y_1}}}x + {1 \over {4{y_1}}}

$\Rightarrow$ 4y1y = x1x + 1 which intersects x axis at A

\left( {{{ - 1} \over {{x_1}}},0} \right)

and y axis at

B\left( {0,{1 \over {4{y_1}}}} \right)

Let midpoint of AB is (h, k) $\therefore$ h =

{{ - 1} \over {2{x_1}}}

$\Rightarrow$

\,\,\,

x1 =

{{ - 1} \over {2h}}

& y1 =

{1 \over {8k}}

Thus, 4

{\left( {{1 \over {8k}}} \right)^2}

=

{\left( {{{ - 1} \over {2h}}} \right)^2}

+ 1 $\Rightarrow$

\,\,\,

{1 \over {16{k^2}}}

=

{1 \over {4{h^2}}}

+ 1 $\Rightarrow$

\,\,\,

1 =

{{16{k^2}} \over {4{h^2}}}

+ 16k2 $\Rightarrow$

\,\,\,

h2 = 4k2 + 16h2 k. So, required equation is x2 $-$ 4y2 $-$ 16x2 y2 = 0

Q54

A normal to the hyperbola, 4x2

-

9y2 = 36 meets the co-ordinate axes

x

and y at A and B, respectively. If the parallelogram OABP (O being the origin) is formed, then the ocus of P is :

A 4x2 + 9y2 = 121

B 9x2 + 4y2 = 169

C 4x2

-

9y2 = 121

D 9x2

-

4y2 = 169

Correct Answer

Option D

Solution

Given, 4x2 $-$ 9y2 = 36 After differentiating w.r.t.x, we get 4.2x $-$ 9.2.y.

{{dy} \over {dx}}

= 0 $\Rightarrow$ Slope of tangent =

{{dy} \over {dx}}

=

{{4x} \over {9y}}

So, slope of normal =

{{ - 9y} \over {4x}}

Now, equation of normal at point (x0, y0) is given by y $-$ y0 =

{{ - 9{y_0}} \over {4{x_0}}}

(x $-$ x0) As normal intersects X axis at A, Then A =

\left( {{{13{x_0}} \over 9},0} \right)

and B $\equiv$

\left( {0,{{13{y_0}} \over 4}} \right)

As OABP is parallelogram $\therefore$ midpoint of OB $\equiv$

\left( {0,{{13{y_0}} \over 8}} \right)

$\equiv$ Midpoint of AP So, P(x, y) $\equiv$

\left( {{{ - 13{x_0}} \over 9},{{13{y_0}} \over 4}} \right)

...(i) $\because$ (x0, y0) lies on hyperbola, therefore 4(x0)2 $-$ 9(y0)2 = 36 From equation (i) : x0 =

{{ - 9x} \over {13}}

and y0 =

{{4y} \over {13}}

From equation (ii), we get 9x2 $-$ 4y2 = 169 Hence, locus of point P is : 9x2 $-$ 4y2 = 169

Q55

The locus of the point of intersection of the lines,

\sqrt 2 x - y + 4\sqrt 2 k = 0

and

\sqrt 2 k\,x + k\,y - 4\sqrt 2 = 0

(k is any non-zero real parameter), is :

A an ellipse whose eccentricity is

{1 \over {\sqrt 3 }}.

B an ellipse with length of its major axis

8\sqrt 2 .

C a hyperbola whose eccentricity is

\sqrt 3 .

D a hyperbola with length of its transverse axis

8\sqrt 2 .

Correct Answer

Option D

Solution

Here, lines are :

\sqrt 2 x

$-$ y + 4

\sqrt 2 k

= 0 $\Rightarrow$

\,\,\,

\sqrt 2 x + 4\sqrt 2 k = y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,....

(i) and

\sqrt 2 kx + ky - 4\sqrt 2 = 0\,\,\,\,\,...\left( {ii} \right)

Put the value of y from (i) in (ii) we get; $\Rightarrow$ 2

\sqrt 2

kx + 4

\sqrt 2

(k2 $-$ 1) = 0 $\Rightarrow$ x =

{{2\left( {1 - {k^2}} \right)} \over k}

, y =

{{2\sqrt 2 \left( {1 + {k^2}} \right)} \over k}

\therefore\,\,\,

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

\therefore\,\,\,

length of transverse axis 2a = 2 $\times$ 4

{\sqrt 2 }

= 8

{\sqrt 2 }

Hence, the locus is a hyperbola with length of its transverse axis equal to 8

{\sqrt 2 }

Q56

Equation of a common tangent to the parabola y2 = 4x and the hyperbola xy = 2 is :

A x + y + 1 = 0

B 4x + 2y + 1 = 0

C x – 2y + 4 = 0

D x + 2y + 4 = 0

Correct Answer

Option D

Solution

Let the equation of tangent to parabola y2 = 4x be y = mx +

{1 \over m}

It is also a tangent to hyperbola xy = 2 $\Rightarrow$ x

\left( {mx + {1 \over m}} \right)

= 2 $\Rightarrow$ x2m +

{x \over m}

$-$ 2 = 0 D = 0 $\Rightarrow$ m = $-$

{1 \over 2}

So tangent is 2y + x + 4 = 0

Q57

The equation of a tangent to the hyperbola 4x2 – 5y2 = 20 parallel to the line x – y = 2 is :

A x

-

y + 9 = 0

B x

-

y

-

3 = 0

C x

-

y + 1 = 0

D x

-

y + 7 = 0

Correct Answer

Option C

Solution

Hyperbola

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

slope of tangent = 1 equation of tangent y = x $\pm$

\sqrt {5 - 4}

$\Rightarrow$ y = x $\pm$ 1 $\Rightarrow$ y = x + 1 or $\Rightarrow$ y = x $-$ 1

Q58

A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is :

A

{3 \over 2}

B

\sqrt 3

C 2

D

{2 \over {\sqrt 3 }}

Correct Answer

Option D

Solution

Let the equation of hyperbola

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

= 1 Given 2a = 4 $\Rightarrow$

a

= 2 It passes through (4, 2) $\therefore$

{{16} \over 4} - {4 \over {{b^2}}}

= 1 $\Rightarrow$ b2 =

{4 \over 3}

e =

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

=

\sqrt {1 + {{4/3} \over 4}}

=

\sqrt {1 + {1 \over 3}}

=

{2 \over {\sqrt 3 }}

Q59

Let

0 < \theta < {\pi \over 2}

. If the eccentricity of the hyperbola

{{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }}

= 1 is greater than 2, then the length of its latus rectum lies in the interval :

A (3,

\infty

)

B

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

C

\left( {1,{3 \over 2}} \right]

D

\left( {2,3} \right]

Correct Answer

Option A

Solution

Given hyperbola,

{{{x^2}} \over {{{\cos }^2}\theta }} - {{{y^2}} \over {{{\sin }^2}\theta }} = 1

here a = cos $\theta$ and b = sin $\theta$ We know, eccentricity of the hyperbola is,

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

$\therefore$ Here eccentricity (e) =

\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}}

Given that,

\sqrt {1 + {{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}} > 2

$\Rightarrow$

\sqrt {1 + {{\tan }^2}\theta } > 2

$\Rightarrow$ 1 + tan2 $\theta$ > 4 $\Rightarrow$ tan2 $\theta$ > 3 $\Rightarrow$ tan $\theta$ >

\pm \sqrt 3

As given $\theta$

\in

\left( {0,{\pi \over 2}} \right)

possible value of tan $\theta$ >

\sqrt 3

So, $\theta$ can be in the range

{\pi \over 3} < \theta < {\pi \over 2}

We know latus ractum (LR) =

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

$\therefore$ LR =

{{2{{\sin }^2}\theta } \over {\cos \theta }}

= 2 tan $\theta$ sin $\theta$ We know in the range

{\pi \over 3} < \theta < {\pi \over 2}

tan $\theta$ and sin $\theta$ both are increasing function. So, at

{\pi \over 3}

value of LR will be minimum and at

{\pi \over 2}

value of LR will be maximum. $\therefore$ Minimum value of LR = 2tan

{\pi \over 3}

sin

{\pi \over 3}

= 2 $\times$

\sqrt 3 \times {{\sqrt 3 } \over 2}

= 3 Maximum value of LR = 2tan

{\pi \over 2}

sin

{\pi \over 2}

= 2

\left( \infty \right) \times 1

= $\infty$ $\therefore$ Interval of LR = (3, $\infty$ )

Q60

A circle cuts a chord of length 4a on the x-axis and passes through a point on the y-axis, distant 2b from the origin. Then the locus of the centre of this circle, is :

A an ellipse

B a parabola

C a hyperbola

D a straight line

Correct Answer

Option B

Solution

Let equation of circle is x2 + y2 + 2fx + 2fy + e = 0, it passes through (0, 2b) $\Rightarrow$ 0 + 4b2 + 2g $\times$ 0 + 4f + c = 0 $\Rightarrow$ 4b2 + 4f + c = 0 . . . (i)

2\sqrt {{g^2} - c} = 4a

. . . (ii) g2 $-$ c = 4a2 $\Rightarrow$ c =

\left( {{g^2} - 4{a^2}} \right)

Putting in equation (1) $\Rightarrow$ 4b2 + 4f + g2 $-$ 4a2 = 0 $\Rightarrow$ x2 + 4y + 4(b2 $-$ a2) = 0, it represent a hyperbola.