Hyperbola

Let focal chord be latusrectum then it is given that it subtends 90$^\circ$ at its centre.

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Here, equation of hyperbola is

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ ...... (i)

So eccentricity, $e=\sqrt{1+\frac{b^2}{a^2}}$

End points of latus rectum, $L=\left(a e, \frac{b^2}{a}\right)$ and $L^{\prime}\left(a e, \frac{-b^2}{a}\right) $

Centre $(0,0)$

$\because < L C L^{\prime}=90^{\circ}$

In $\triangle L C L^{\prime}$ by Pythagoras theorem

$\begin{aligned} & (L C)^2+\left(L^{\prime} C\right)^2=\left(L L^{\prime}\right)^2 \\ & (a e-0)^2+\left(\frac{b^2}{a^2}-0\right)^2+(a e-0)^2+\left(-\frac{b^2}{a}-0\right)^2=\left(\frac{2 b^2}{a}\right)^2 \\ & \Rightarrow \quad 2\left(a^2 e^2+\frac{b^4}{a^2}\right)=\frac{4 b^4}{a^2} \Rightarrow b^4=a^4 e^2 \\ & \Rightarrow \quad\left(e^2-1\right)^2=e^2 \quad \quad \quad\left[\because b^2=a^2\left(e^2-1\right)\right] \\ & \Rightarrow \quad {e^2 \pm e-1}{}=0 \\ & \Rightarrow e=\pm \frac{1\pm \sqrt{1+4}}{2}=\frac{\pm1\pm\sqrt5}{2}\end{aligned}$

$\because e > 1$ for hyperbola

$\Rightarrow e=\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2} \Rightarrow e=\frac{\sqrt{5}+1}{2}$

$\Rightarrow$ Option (c) is correct.

Distance from focus to directrix is $a\left(e-\frac{1}{e}\right)$

$a\left(\frac{5}{4}-\frac{4}{5}\right)=\frac{4 \cdot 3-3}{|5|} \Rightarrow a=4$

Slope of axis $=(-3 / 4)$

Distance from vertex to focus

$\begin{aligned} & =a e-a=a\left(\frac{5}{4}\right)-a=\frac{a}{4} \\ \text { Vertex } & =\left[3 \pm\left(\frac{-4}{5}\right), 0 \pm\left(\frac{3}{5}\right)\right] \\ & =\left(3-\frac{4}{5}, \frac{3}{5}\right)=\left(\frac{11}{5}, \frac{3}{5}\right) \end{aligned}$

Any tangent to hyperbola forms a triangle with asymptotes which has constant area $=a b$

$\begin{aligned} & \Rightarrow \quad a b=a^2 \tan \alpha \Rightarrow \frac{b}{a}=\tan \alpha \\ & \text { Eccentricity, } e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{1+\tan ^2 \alpha}=\sec \alpha \end{aligned}$

x = 9 meets the hyperbola at (9, 6$\sqrt2$) and (9, $-$6$\sqrt2$). Then, the equations of tangent at these points are 3x $-$ 2$\sqrt{2y}$ $-$ 3 = 0. The combined equation of these two tangent is 9x² $-$ 8y² $-$ 18x + 9 = 0

Let assume equation of hyperbola

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

Given that length of transverse axis, $2 a=12$ or $a=6$

∵ Point $(8,2)$ lies on hyperbola, then it must satisfy the equation of hyperbola

$ \begin{aligned} i.e.,\frac{(8)^2}{a^2}-\frac{(2)^2}{b^2} & =1 \text { or } \frac{64}{a^2}-\frac{4}{b^2}=1 \\ \frac{64}{(6)^2}-\frac{4}{b^2} & =1 \text { or } \frac{4}{b^2}=\frac{16}{9}-1 \\ & =\frac{7}{9} \text { or } b^2=\frac{36}{7} \end{aligned} $

As we know eccentricity of parabola,

$ e=\frac{\sqrt{a^2+b^2}}{a}=\frac{\sqrt{(6)^2+\frac{36}{7}}}{6}=\frac{2 \sqrt{2}}{\sqrt{7}} $

$p$ and $q$ are eccentricity of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate of hyperbola.

$ \begin{align*} & \therefore p^2=\frac{a^2+b^2}{a^2} \text { and } q^2=\frac{a^2+b^2}{b^2} \\ & \quad \frac{x^2}{p^2}+\frac{y^2}{q^2}=1 \\ & \Rightarrow a^2 x^2+b^2 y^2=a^2+b^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

and pair of line, $x^2-y^2=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Solving Eqs. (i) and (ii), we get

$ (1,1),(-1,-1),(-1,1)(1,-1) $

$ \begin{aligned} &\text { Area of square }\\ &A B C D,=A B^2=(2)^2=4 \end{aligned} $

We have,

$ \begin{aligned} x^2+y^2 & =a^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\and \,\,\,\,\,x y & =b^2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Let four points

$A\left(x_1, y_1\right), B\left(x_2, y_2\right), C\left(x_3, y_3\right)$ and $D\left(x_4, y_4\right)$

$ \begin{array}{ll} \therefore & O A^2=x_1^2+y_1^2 \\ & O B^2=x_2^2+y_2^2 \\ & O C^2=x_3^2+y_3^2 \\ \text { and } & O D^2=x_4^2+y_4^2 \end{array} $

Such that $O A^2+O B^2+O C^2+O D^2$

$ =\Sigma x_1^2+\Sigma y_1^2=4 a^2 $

From Eq. (ii) $x=\frac{b^2}{y}$

From Eq. (i) $\left(\frac{b^2}{y}\right)^2+y^2=a^2$

$ \begin{aligned} & \Rightarrow & b^4+y^4 & =a^2 y^2 \\ & \Rightarrow & y^4-a^2 y^2+b^4 & =0 \end{aligned} $

This is an equation of 4 th degree in $y$ and its four roots are $y_1, y_2, y_3, y_4$, then $y_1 y_2 y_3 y_4=b^4$

Let the equation of hyperbola be

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

We have,

$ \begin{aligned} & \quad e=\sqrt{2} \text { and } 2 a e=16 \Rightarrow 2 a(\sqrt{2})=16 \\ & \Rightarrow \quad a=\frac{16}{2 \sqrt{2}}=4 \sqrt{2} \end{aligned} $

Now, $e=\sqrt{2} \Rightarrow e^2=2$

$ \begin{aligned} & \Rightarrow 1+\frac{b^2}{a^2}=2 \Rightarrow \frac{b^2}{a^2}=1 \\ & \Rightarrow \quad b^2=a^2 \Rightarrow b=a=4 \sqrt{2} \end{aligned} $

∴ Equation of hyperbola is

$ \frac{x^2}{32}-\frac{y^2}{32}=1 \Rightarrow x^2-y^2=32 $

$\bullet$ For the given equation $a = \sqrt 2 $ and point ($-$1, 1), x² + y² = a² satisfies these conditions (i.e. I of Column 1)

$\bullet$ Thus, equation of tangent is $ - x + y = 2 \Rightarrow y = x + 2$; thus (ii) of Column 2 satisfies this condition with m = 1. That is, $y = mx + a\sqrt {{m^2} + 1} $

$\bullet$ It is given that point of constant ($-$1, 1) given m = 1 and $a = \sqrt 2 $ Q; of Columns 3 : $\left( {{{ - ma} \over {\sqrt {{m^2} + 1} }},{a \over {\sqrt {{m^2} + 1} }}} \right)$ gives point of contact ($-$1, 1).

Thus, correct combination is (I) (II) (Q); thus, option (A) is correct.

Given, tangent to conic is $\sqrt 3 x + 2y = 4$ and point of contact is

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

$\sqrt 3 x + 2y = 4 \Rightarrow 2y = 4 - \sqrt 3 x$

$ \Rightarrow y = {{ - \sqrt 3 } \over 2}x + 2$ ....... (2)

On comparing this equation with options in Column 2, we get

$m = {{ - \sqrt 3 } \over 2}$

Equation (1) can also be written as

$ \Rightarrow \sqrt 3 x + 4\left( {{1 \over 2}} \right)y = 4$

On comparing with the option (II), we get

${a^2} = 4 \Rightarrow a = 2$

Now, for a = 2, $m = {{ - \sqrt 3 } \over 2}\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} + 1} }},{1 \over {\sqrt {{a^2}{m^2} + 1} }}} \right)$ gives $\left( {\sqrt 3 ,{1 \over 2}} \right)$ as the point of contact.

Thus, the correct combination is (II) (iv) (R); hence, option (B) is correct.

Line y = mx + C is tangent to hyperbola ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$ if following condition is satisfied:

${C^2} = {a^2}{m^2} - {b^2}$ ..... (1)

Given : The equation of line = $2x - y + 1 = 0$

Hyperbola = ${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{{16}^2}}} = 1$

Consider equation $2x - y + 1 = 0$

$ \Rightarrow y = 2x + 1 \Rightarrow m = 2$ and C = 1

From equation of hyperbola, we get b² = 16 and a²

Substituting all values in Eq. (1), we get

${1^2} = {a^2}{(2)^2} - 16$

$1 = 4{a^2} - 16 \Rightarrow 4{a^2} = 16 + 1 \Rightarrow 4{a^2} = 17$

$ \Rightarrow {a^2} = {{17} \over 4} \Rightarrow a = \sqrt {{{17} \over 4}} = {{\sqrt {17} } \over 2}$

Therefore, $a = {{\sqrt {17} } \over 2}$

Option (A) : a, 4, 1 $ \Rightarrow {{\sqrt {17} } \over 2}$, 4, 1 is not a right triangle (Since $\sqrt {{4^2} + {1^1}} \ne {{\sqrt {17} } \over 2}$)

Option (B) : 2a, 4, 1 $\sqrt {17} $, 4, 1 is a right angled triangle (since $\sqrt {{4^2} + {1^2}} = \sqrt {16 + 1} = \sqrt {17} $)

Option (C) : a, 4, 2 $ \Rightarrow {{\sqrt {17} } \over 2}$, 4, 2 is not a right angled triangle (Since $\sqrt {{4^2} + {2^2}} \ne {{\sqrt {17} } \over 2}$)

Option (D) : 2a, 8, 1 $\sqrt {17} $, 8, 1 is not a right angled triangle (Since $\sqrt {{8^2} + {1^2}} \ne \sqrt {17} $)

Equation of family of circles touching hyperbola at (x₁, y₁) is

(x $-$ x₁)² + (y $-$ y₁)² + $\lambda$(xx₁ $-$ yy₁ $-$ 1) =0

Now, its centre is (x₂, 0).

$\therefore$ $\left[ {{{ - (\lambda {x_1} - 2{x_1})} \over 2},{{ - ( - 2{y_1} - \lambda {y_1})} \over 2}} \right] = ({x_2},0)$

$ \Rightarrow 2{y_1} + \lambda {y_1} = 0 \Rightarrow \lambda = - 2$

and $2{x_1} - \lambda x = 2{x_2} \Rightarrow {x_2} = 2{x_1}$

$\therefore$ $P({x_1},\sqrt {x_1^2 - 1} )$

and $N({x_2},0) = (2{x_1},0)$

As tangent intersect X-axis at $M\left( {{1 \over x},0} \right)$.

Centroid of $\Delta PWN = (l,m)$

$ \Rightarrow \left( {{{3{x_1} + {1 \over {{x_1}}}} \over 3},{{{y_1} + 0 + 0} \over 3}} \right) = (l,m)$

$ \Rightarrow l = {{3{x_1} + {1 \over {{x_1}}}} \over 3}$

On differentiating w.r.t. x₁, we get

${{dl} \over {d{x_1}}} = {{3 - {1 \over {x_1^2}}} \over 3}$

$ \Rightarrow {{dl} \over {d{x_1}}} = 1 - {1 \over {3x_2^1}}$, for x₁ > 1

and $m = {{\sqrt {x_1^2 - 1} } \over 3}$

On differentiating w.r.t. x₁, we get

${{dm} \over {d{x_1}}} = {{2{x_1}} \over {2 \times 3\sqrt {x_1^2 - 1} }} = {{{x_1}} \over {3\sqrt {x_1^2 - 1} }}$, for x₁ > 1

Also, $m = {{{y_1}} \over 3}$

On differentiating w.r.t. y₁, we get

${{dm} \over {d{y_1}}} = {1 \over 3}$, for y₁ > 0

Since, the tangents are parallel to the straight line $2 x-y=1$.

$\therefore$ Equation of the tangents is

$\Rightarrow \quad \begin{aligned} 2 x-y+\mathrm{C} & =0 \\ y & =2 x+\mathrm{C} \end{aligned}$

$\Rightarrow$ Slope of the tangents is $m=2$.

Now, from the standard equation of given hyperbola.

$\begin{aligned} & \quad a^2=9, b^2=4 \\ \therefore \quad & C^2=a^2 m^2-b^2 \\ \Rightarrow \quad & C^2=9 \times 2^2-4 \\ \Rightarrow \quad & C^2=32 \\ \Rightarrow \quad & C= \pm 4 \sqrt{2} \end{aligned}$

$\therefore$ Point of contact is $\left( \pm \frac{a^2 m}{\mathrm{C}}, \pm \frac{b^2}{\mathrm{C}}\right)$

i.e. $\left( \pm \frac{9 \times 2}{4 \sqrt{2}}, \pm \frac{4}{4 \sqrt{2}}\right)$

i.e. $\left( \pm \frac{9}{2 \sqrt{2}}, \pm \frac{1}{\sqrt{2}}\right)$

Equation of normal is

$(y - 3) = {{ - {a^2}} \over {2{b^2}}}(x - 6) \Rightarrow {{{a^2}} \over {2{b^2}}} = 1 \Rightarrow e = \sqrt {{3 \over 2}} $

Ellipse is ${{{x^2}} \over {{2^2}}} + {{{y^2}} \over {{1^2}}} = 1$.

${1^2} = {2^2}(1 - {e^2}) \Rightarrow e = {{\sqrt 3 } \over 2}$

Therefore, the eccentricity of the hyperbola is

${2 \over {\sqrt 3 }} \Rightarrow {b^2} = {a^2}\left( {{4 \over 3} - 1} \right) \Rightarrow 3{b^2} = {a^2}$

Foci of the ellipse are $(\sqrt 3 ,\,0)$ and $( - \sqrt 3 ,\,0)$.

Hyperbola passes through $( \sqrt 3 ,\,0)$.

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

Therefore, the equation of hyperbola is ${x^2} - 3{y^2} = 3$.

Focus of hyperbola is $(ae,\,0) \equiv \left( {\sqrt 3 \times {2 \over {\sqrt 3 }},\,0} \right) \equiv (2,0)$.

A point on hyperbola is (3sec$\theta$, 2tan$\theta$).

It lies on the circle, so

$9{\sec ^2}\theta + 4{\tan ^2}\theta - 24\sec \theta = 0$.

$ \Rightarrow 13{\sec ^2}\theta - 24\sec \theta - 4 = 0 \Rightarrow \sec \theta = 2, - {2 \over {13}}$

Therefore, $\sec \theta = 2 \Rightarrow \tan \theta = \sqrt 3 $

The point of intersection are $A(6,2\sqrt 3 )$ and $B(6, - 2\sqrt 3 )$

Hence, The circle with AB as diameter is

${(x - 6)^2} + {y^2} = {(2\sqrt 3 )^2} \Rightarrow {x^2} + {y^2} - 12x + 24 = 0$

A tangent to ${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$ is

$y = mx + \sqrt {9{m^2} - 4} ,\,m > 0$ .... (1)

A tangent to ${(x - 4)^2} + {y^2} = 16$ is

$xy = m(x - 4) + 4\sqrt {1 + {m^2}} $ ..... (2)

Comparing (1) and (2),

$\sqrt {9{m^2} - 4} = - 4m + 4\sqrt {1 + {m^2}} \Rightarrow \sqrt {9 - {4 \over {{m^2}}}} = - 4 + 4\sqrt {1 + {1 \over {{m^2}}}} $

Let ${1 \over {{m^2}}} = t$, we have $\sqrt {9 - 4t} = - 4 + 4\sqrt {1 + t} $

Squaring, we have

$ \Rightarrow 9 - 4t = 16 + 16(1 + t) - 32\sqrt {1 + t} \Rightarrow 32\sqrt {1 + t} = 23 + 20t$

Again squaring $1024(1 + t) = 529 + 920t + 400{t^2}$

$ \Rightarrow 400{t^2} - 104t - 495 = 0 \Rightarrow t = {5 \over 4}$

Thus ${m^2} = {4 \over 5},\,m = {2 \over {\sqrt 5 }}$

The tangent is $y = {2 \over {\sqrt 5 }}x + {4 \over {\sqrt 5 }}$ i.e. $2x - \sqrt 5 y + 4 = 0$