Hyperbola

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. $2x - \sqrt 5 y + 4 = 0$

We have,

${x^2} - 2{y^2} - 2\sqrt 2 x - 4\sqrt 2 y - 6 = 0$

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

$a = 2,b = \sqrt 2 $

$ \Rightarrow e = \sqrt {{3 \over 2}} $

Now, $e = \sqrt {{{{a^2} + {b^2}} \over {{a^2}}}} = \sqrt {{3 \over 2}} $

Area, $ = {1 \over 2}$ $\times$ Base $\times$ Height

$ = {1 \over 2}a(e - 1){{{b^2}} \over a}$

$ = {1 \over 2}{{(\sqrt 3 - \sqrt 2 ) \times 2} \over {\sqrt 2 }}$

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

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

The length of transverse axis $ = 2\sin \theta = 2a$

$a = \sin \theta $

also for ellipse $3{x^2} + 4{y^2} = 12$

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

${a^2} = 4,{b^3} = 3$

$e = \sqrt {1 - {{{b^2}} \over {{a^2}}}} $

$\therefore$ $e = \sqrt {1 - {3 \over 4}} = {1 \over 2}$

$\therefore$ Focus of ellipse $ = \left( {2 \times {1 \over 2},0} \right) \Rightarrow (1,0)$

As hyperbola is confocal with ellipse, focus of hyperbola is = (1, 0)

$ae = 1 \Rightarrow \sin \theta \times e = 1$

$e = \cos ec\theta $

$\therefore$ ${b^2} = {a^2}({e^2} - 1)$

${\sin ^2}\theta (\cos e{c^2}\theta - 1) = {\cos ^2}\theta $

$\therefore$ equation of hyperbola is

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

${x^2}\cos e{c^2}\theta - {y^2}{\sec ^2}\theta = 1$

The equation of the ellipse is $\frac{x^2}{25}+\frac{y^2}{16}=1$.

Here, the numbers 25 and 16 are the squares of the lengths of the axes. This means the semi-major axis $ a = 5 $ and the semi-minor axis $ b = 4 $.

Step 1: Find the eccentricity of the ellipse.

The eccentricity of an ellipse is given by the formula: $e = \sqrt{1 - \frac{b^2}{a^2}}$

Let’s plug in the values: $ b^2 = 16 $, $ a^2 = 25 $

So, $ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} $

Step 2: Find the focus of the ellipse.

The foci are at $ (\pm \sqrt{a^2 - b^2}, 0) $.

So, $ a^2 - b^2 = 25 - 16 = 9 $. Take square root: $ \sqrt{9} = 3 $.

This means the foci are at $ (\pm 3, 0) $.

Step 3: Eccentricity of the hyperbola.

The problem tells us: Product of eccentricities is 1.

The ellipse's eccentricity $ e_{ellipse} = \frac{3}{5} $. Let $ e_{hyperbola} $ be the hyperbola's eccentricity.

So, $ e_{hyperbola} \times \frac{3}{5} = 1 $. $ e_{hyperbola} = \frac{1}{\frac{3}{5}} = \frac{5}{3} $

Step 4: Hyperbola passes through the focus (3, 0).

Since the axes of the hyperbola match the ellipse, and it passes through (3, 0), write: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Plug in $ x = 3 $, $ y = 0 $:

$\frac{3^2}{a^2} - 0 = 1 \implies \frac{9}{a^2} = 1 \implies a^2 = 9$

So the hyperbola equation is: $\frac{x^2}{9} - \frac{y^2}{b^2} = 1$

Step 5: Use the eccentricity formula for the hyperbola.

For a hyperbola, eccentricity $ e_{hyperbola} = \sqrt{1 + \frac{b^2}{a^2}} $

We know $ e_{hyperbola} = \frac{5}{3} $, $ a^2 = 9 $. Set up the equation:

$\sqrt{1 + \frac{b^2}{9}} = \frac{5}{3}$

Square both sides: $1 + \frac{b^2}{9} = \frac{25}{9}$

So, $ b^2 = 16 $.

Step 6: Final equation of the hyperbola.

So, the hyperbola is $\frac{x^2}{9} - \frac{y^2}{16} = 1$ with $ a^2 = 9, b^2 = 16 $.

Step 7: Find the foci of the hyperbola.

The foci are at $ (\pm \sqrt{a^2 + b^2}, 0) $.

Plug in $ a^2 = 9, b^2 = 16 $: $ \sqrt{9 + 16} = \sqrt{25} = 5 $.

So, the foci are at $ (\pm 5, 0) $.