Hyperbola

Equation of given hyperbola is

${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$, a > b and a > 1

Let point P(a sec$\theta $, b tan$\theta $) on the hyperbola in first quadrant i.e., $Qe\left( {0,{\pi \over 2}} \right)$.

Now equation of tangent to the hyperbola at point P is

${{x\sec \theta } \over a} - {{y\tan \theta } \over b} = 1$ .... (i)

$ \because $ The tangent (i) passes through point A(1, 0)

So, $a = \sec \theta > 1$. ....(ii)

and equation of normal to the hyperbola at point P having slope is ${{b\sec \theta } \over {a\tan \theta }}$, as normal cuts off equal intercepts on the coordinate axes, so slope must be $-$1.

Therefore, ${{b\sec \theta } \over {a\tan \theta }}$ = $-$1

$ \Rightarrow $ $b = - \tan \theta $ {from Eq. (ii)}

$ \because $ The eccentricity of hyperbola

$e = \sqrt {1 + {{{b^2}} \over {{a^2}}}} = \sqrt {1 + {{\sin }^2}\theta } $

$ \therefore $ $1 < e < \sqrt 2 $, as $\theta \in \left( {0,{\pi \over 2}} \right)$

And the area of required triangle is, which is isosceles is

$\Delta = {1 \over 2}{(AP)^2} = {1 \over 2}[{(1 - {\sec ^2}\theta )^2} + {\tan ^4}\theta ]$



$ = {1 \over 2}(2{\tan ^4}\theta ) = {\tan ^4}\theta = {b^4}$.

It is given that T is tangents to S₁ at P and S₂ at Q and S₁ and S₂ touch externally at M.



$ \therefore $ MN = NP = NQ

$ \therefore $ Locus of M is a circle having PQ as its diameter of circle

$ \therefore $ Equation of circle

$(x - 2)(x + 2) + (y + 5)(y - 7) = 0$

$ \Rightarrow {x^2} + {y^2} - 2y - 39 = 0$

Hence,

${E_1}:{x^2} + {y^2} - 2y - 39 = 0$, $x \ne \pm 2$

Locus of mid-point of chord (h, k) of the circle E₁ is

$xh + yk - (y + k) - 39 = {h^2} + {k^2} - 2k - 39$

$ \Rightarrow xh + yk - y - k = {h^2} + {k^2} - 2k$

Since, chord is passing through (1, 1).

$ \therefore $ Locus of mid-point of chord (h, k) is

$h + k - 1 - k = {h^2} + {k^2} - 2k$

$ \Rightarrow {h^2} + {k^2} - 2k - h + 1 = 0$

Locus is ${E_2}:{x^2} + {y^2} - x - 2y + 1 = 0$

Now, after checking options, (a) and (d) are 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. $\left( \pm \frac{9}{2 \sqrt{2}}, \pm \frac{1}{\sqrt{2}}\right)$

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

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) $.