Hyperbola

To transform the equation $x^2 - y^2 + 2x + 4y = 0$ when the origin is shifted to the point $(-1, 2)$, we proceed as follows:

Given equation:

$ x^2 - y^2 + 2x + 4y = 0 $

New origin:

$ (h, k) = (-1, 2) $

Transform the variables:

Let $ X = x - h $ and $ Y = y - k $.

So,

$ X = x - (-1) = x + 1 $

$ Y = y - 2 $

Solving for $ x $ and $ y $:

$ x = X - 1 $

$ y = Y + 2 $

Now, substituting these into the given equation:

$ (x - 1)^2 - (y + 2)^2 + 2(x - 1) + 4(y + 2) = 0 $

Expanding each term:

$(x - 1)^2 = x^2 - 2x + 1$

$(y + 2)^2 = y^2 + 4y + 4$

Substitute and simplify:

$ \begin{aligned} & [x^2 - 2x + 1] - [y^2 + 4y + 4] + 2(x - 1) + 4(y + 2) = 0 \\ & x^2 - y^2 - 2x + 1 - 4y - 4 + 2x - 2 + 4y + 8 = 0 \end{aligned} $

Combine and simplify:

$ x^2 - y^2 + 3 = 0 $

Thus, the transformed equation is:

$ x^2 - y^2 + 3 = 0 $

Given the ellipse equation:

$ 4x^2 + 9y^2 = 36 $

or equivalently,

$ \frac{x^2}{9} + \frac{y^2}{4} = 1 \quad \text{... (i)} $

This ellipse is confocal with a hyperbola whose transverse axis has a length of 2. For the hyperbola, the transverse axis is $2a = 2$, giving us $a = 1$.

For confocal conics, the foci are at the same distance from the origin. The distance of the focus for the ellipse is:

$ c = \sqrt{a^2 - b^2} = \sqrt{\sqrt{9^2} - \sqrt{4^2}} = \sqrt{9 - 4} = \sqrt{5} $

For the hyperbola, we have:

$ c = \sqrt{a^2 + b^2} $

Thus,

$ c = \sqrt{1 + b^2} = \sqrt{5} $

Solving for $b^2$, we find $b^2 = 4$. Therefore, the equation of the hyperbola is:

$ \frac{x^2}{1} - \frac{y^2}{4} = 1 \quad \Rightarrow \quad x^2 - \frac{y^2}{4} = 1 \quad \text{... (ii)} $

This implies:

$ y^2 = 4(x^2 - 1) \quad \text{... (iii)} $

From equations (i) and (iii), we get:

$ \begin{align*} 4x^2 + 9\left(4(x^2 - 1)\right) &= 36 \\ 4x^2 + 36x^2 - 36 &= 36 \\ 40x^2 &= 72 \\ x^2 &= \frac{9}{5} \\ x &= \pm \frac{3\sqrt{5}}{5} \end{align*} $

Thus, using equation (iii):

$ \begin{align*} y^2 &= 4 \times \left(\frac{9}{5} - 1\right) = 4 \times \frac{4}{5} = \frac{16}{5} \\ y &= \pm \frac{4\sqrt{5}}{5} \end{align*} $

The points of intersection are $\left( \pm \frac{3\sqrt{5}}{5}, \frac{4\sqrt{5}}{5} \right)$.

The equation of a circle passing through these points is:

$ \begin{align*} x^2 + y^2 &= \left(\frac{3\sqrt{5}}{5}\right)^2 + \left(\frac{4\sqrt{5}}{5}\right)^2 \\ &= \frac{45 + 80}{25} = \frac{125}{25} \\ \therefore \quad x^2 + y^2 &= 5 \end{align*} $

For the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, the eccentricity $e$ is given by the formula:

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

Since the eccentricity is provided as $\sec \alpha$, we have:

$ \sec \alpha = \sqrt{1+\frac{b^2}{a^2}} $

Squaring both sides of the equation, we obtain:

$ \sec^2 \alpha = 1 + \frac{b^2}{a^2} $

Subtracting 1 from both sides gives:

$ \sec^2 \alpha - 1 = \frac{b^2}{a^2} = \tan^2 \alpha $

or

$ \frac{b^2}{a^2} = \tan^2 \alpha $

The equations of the asymptotes of the hyperbola are:

$ y = \pm \frac{b}{a} x = \pm \tan \alpha \cdot x $

A general form for the tangent to the hyperbola is:

$ \frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1 $

Given $x_1 = a \sec \theta$ and $y_1 = b \tan \theta$, when $x=0$, the equation transforms to:

$ \frac{-y \tan \theta}{a \tan \alpha} = 1 \Rightarrow y = -\frac{a \tan \alpha}{\tan \theta} $

Thus, the vertices of the triangle formed by asymptotes and the tangent will be at $(\pm a, 0)$ and $\left(0, -\frac{a \tan \alpha}{\tan \theta}\right)$.

The area of the triangle, with the base as $2a$ and height $-\frac{a \tan \alpha}{\tan \theta}$, is calculated by:

$ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 2a \times \left(\frac{a \tan \alpha}{\tan \theta}\right) $

Simplifying this, we get:

$ = \left|\frac{a^2 \tan \alpha}{\tan \theta}\right| = \left|\frac{a^2 b^2}{a^2 \tan \theta}\right| $

From our earlier identity, this equals:

$ \text{Area} = \frac{b^2}{\tan \theta} = \frac{b^2}{|\tan \alpha|} $

Given the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, its eccentricity $e_1$ is defined by:

$ e_1 = \sqrt{1 + \frac{b^2}{a^2}} $

For the conjugate hyperbola $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$, the eccentricity $e_2$ is:

$ e_2 = \sqrt{1 + \frac{a^2}{b^2}} $

We are given the line:

$ \frac{x}{2e_1} + \frac{y}{2e_2} = 1 $

Rewriting it as:

$ \frac{x}{2e_1} + \frac{y}{2e_2} - 1 = 0 \tag{ii} $

Our task is to find the radius of the circle, centered at the origin, which this line touches.

The distance from the origin $(0,0)$ to a line $Ax + By + C = 0$ is:

$ d = \frac{|C|}{\sqrt{A^2 + B^2}} $

For equation (ii), the distance becomes:

$ d = \frac{|-1|}{\sqrt{\left(\frac{1}{2e_1}\right)^2 + \left(\frac{1}{2e_2}\right)^2}} $

Substituting, we find:

$ d = \frac{2}{\sqrt{\frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2}}} $

$ d = \frac{2}{\sqrt{\frac{a^2 + b^2}{a^2 + b^2}}} = \frac{2}{\sqrt{1}} $

So, the distance $d$ is 2. Therefore, the radius of the circle is 2.

Explanation

Hyperbola A

Given: $2c = 3 \times \frac{2a}{e}$

This implies: $c = \frac{3a}{e}$

Using the relation $c^2 = a^2 + b^2$, we derive:

$ \left(\frac{3a}{e}\right)^2 = a^2 + b^2 $

$ \Rightarrow \frac{9a^2}{e^2} = a^2 + b^2 \Rightarrow 9a^2 = e^2(a^2 + b^2) $

$ \Rightarrow 9 = e^2(1 + \frac{b^2}{a^2}) \Rightarrow 9 = e^2(1 + e^2 - 1) $

$ \Rightarrow e^4 = 9 \Rightarrow e = \sqrt{3} $

Hyperbola B

Let $2a$ be the transverse axis and $2b$ be the conjugate axis, given:

$2a = 2 \times 2b \Rightarrow a = 2b$

Using the relation $c^2 = a^2 + b^2$, we derive:

$ c^2 = (2b)^2 + b^2 \Rightarrow c^2 = 5b^2 \Rightarrow c = \sqrt{5}b $

Therefore:

$ e = \frac{\sqrt{5}b}{2b} = \frac{\sqrt{5}}{2} \quad \left[\because e = \frac{c}{a}\right] $

Hyperbola C

Given the asymptotes equations:

$x + y + 1 = 0 \Rightarrow y = -x - 1$

and $x - y + 3 = 0 \Rightarrow y = x - 3$

The standard form of asymptotes for a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by:

$y = \pm \frac{b}{a}x$

The slopes of asymptotes are $-1$ and $1$, meaning:

$ \frac{b}{a} = \pm 1 \Rightarrow b = a $

Using the relation $c^2 = a^2 + b^2$, we derive:

$ c^2 = a^2 + a^2 \Rightarrow c = \sqrt{2}a $

Therefore:

$ e = \frac{\sqrt{2}a}{a} \Rightarrow e = \sqrt{2} $

Conclusion

In descending order of the magnitudes of the eccentricities, we have $\sqrt{3} > \sqrt{2} > \frac{\sqrt{5}}{2}$. Therefore, the order is $A, C, B$.

To determine the equation of the asymptotes for the hyperbola given by:

$ 4x^2 - 9y^2 - 24x - 36y - 36 = 0 $

we first rearrange and complete the square:

Reorganize the terms:

$ (4x^2 - 24x) - (9y^2 + 36y) - 36 = 0 $

Complete the square for each set of terms:

$ 4(x^2 - 6x) - 9(y^2 + 4y) - 36 = 0 $

Completing the square:

$ 4(x^2 - 6x + 9 - 9) - 9(y^2 + 4y + 4 - 4) - 36 = 0 $

Simplifying:

$ 4(x - 3)^2 - 9(y + 2)^2 - 36 = 0 $

Set the equation in standard hyperbola form:

$ 4(x - 3)^2 - 9(y + 2)^2 = 36 $

$ \frac{(x - 3)^2}{9} - \frac{(y + 2)^2}{4} = 1 $

The asymptotes of a hyperbola in the standard form $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ are given by:

$ (y - k) = \pm \frac{b}{a}(x - h) $

For our hyperbola:

$ h = 3 $, $ k = -2 $

$ a = 3 $, $ b = 2 $

Thus, the equations of the asymptotes are:

$ (y + 2) = \pm \frac{2}{3}(x - 3) $

Expanding this gives:

First Asymptote:

$ y + 2 = \frac{2}{3}(x - 3) $

$ 3(y + 2) = 2(x - 3) $

Simplifying:

$ 3y + 6 = 2x - 6 \qquad \implies \quad 2x - 3y = 12 $

Second Asymptote:

$ y + 2 = -\frac{2}{3}(x - 3) $

$ 3(y + 2) = -2(x - 3) $

Simplifying:

$ 3y + 6 = -2x + 6 \qquad \implies \quad 2x + 3y = 0 $

After multiplying and expanding these, we recover the simplified asymptote equation:

$ 4x^2 - 9y^2 - 24x - 36y = 0 $

The equation of one of the tangent drawn from the point $(0,1)$ to the hyperbola $45 x^2-4 y^2=5$.

We can use the fact that the equation of the tangent to a hyperbola.

$A x^2+B y^2=C$ at point $\left(x_1, y_1\right)$ is given

$A x_1 x+B y_1 y=C$

Here, $A=45, B=-4, C=5$

Let $\left(x_1, y_1\right)$ be the point of tangency, the equation of the tangent line is

$ 45 x_1 x-4 y_1 y=5 $

The line must, also pass through the point ( 0,1 ) substituting ( $x=0$ ) and ( $y=1$ ) into the tangent equation gives

$ \begin{aligned} 45 x_1(0)-4 y_1(1) & =5 \\ \Rightarrow \quad y_1 & =\frac{-5}{4} \\ 45 x_1 x-4\left(\frac{-5}{4}\right) y & =5 \\ 9 x_1 x_1+y & =1 \end{aligned} $

Using the hyperbola equation

$ \begin{aligned} & 45 x_1^2-4\left(\frac{25}{16}\right)=5 \\ \Rightarrow & 45 x_1^2-\frac{25}{4}=5 \\ \Rightarrow & 45 x_1^2=5+\frac{25}{4}=\frac{20+25}{4}=\frac{45}{4} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad 45 x_1^2=\frac{45}{4} \\ & \Rightarrow \quad x_1^2=\frac{1}{4} \Rightarrow x_1= \pm \frac{1}{2} \\ & \Rightarrow \quad x_1=\frac{1}{2} \\ & \Rightarrow \quad 9\left(\frac{1}{2}\right) x+y=1 \\ & \Rightarrow \quad 9 x+2 y=2 \\ & \Rightarrow \quad x_1=\frac{-1}{2} \text { and } 9\left(\frac{-1}{2}\right) x+y=1 \\ & \Rightarrow \quad-9 x+2 y=2 \\ & \Rightarrow 9 x-2 y+2=0 \end{aligned} $

To solve for $ e^2 $ where the hyperbola is given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and the hyperbola passes through the point $(4, -2\sqrt{3})$, we can proceed as follows:

Plug the point into the hyperbola equation:

$ \frac{16}{a^2} - \frac{12}{b^2} = 1 \quad \ldots \text{(i)} $

Relate $ b^2 $ and $ a^2 $ with eccentricity $ e $:

$ b^2 = a^2(e^2 - 1) $

Given the directrix $\sqrt{5}x = 4$:

$ x = \frac{4}{\sqrt{5}} = \frac{a}{e} $

Therefore,

$ a^2 = \frac{16}{5}e^2 \quad \ldots \text{(ii)} $

Substitute the expression for $ a^2 $ from (ii) into (i):

$ \frac{16}{\frac{16}{5}e^2} - \frac{12}{b^2} = 1 $

Substitute $ b^2 = a^2(e^2-1) $:

$ \frac{5}{e^2} - \frac{12}{a^2(e^2 - 1)} = 1 $

Replace $ a^2 $ with $\frac{16}{5}e^2$:

$ \frac{5}{e^2} - \frac{12}{\frac{16}{5}e^2(e^2-1)} = 1 $

Simplify the equation:

$ \frac{5}{e^2} - \frac{60}{16e^2(e^2-1)} = 1 $

On further simplification:

$ \frac{5}{e^2}\left[1 - \frac{12}{16(e^2-1)}\right] = 1 $

Solve for $ e^2 $:

Upon solving, we find that:

$ e^2 = \frac{7}{2} $

This process allows us to determine that the value of $ e^2 $ is $ \frac{7}{2} $.

Given hyperbola $5 x^2-4 y^2-20=0$

$ \begin{array}{ll} \Rightarrow & 5 x^2-4 y^2=20 \\ \Rightarrow & \frac{5 x^2}{20}-\frac{4 y^2}{20}=\frac{20}{20} \\ \Rightarrow & \frac{x^2}{4}-\frac{y^2}{5}=1 \end{array} $

Asymptotes are $x \pm \frac{2}{\sqrt{5}} y=0$

Let $P\left(x_0, y_0\right)$ be any point on $S=0$, Length of the perpendicular from $P$ to $x+\frac{2}{\sqrt{5}} y=0$ is

$ \begin{aligned} l_1 & =\frac{\left.\left|x_0+\frac{2}{\sqrt{5}} y_0\right| \right\rvert\,}{\sqrt{1+\frac{4}{5}}}=\frac{\left|x_0+\frac{2}{\sqrt{5}} y_0\right|}{\sqrt{\frac{9}{5}}} =\frac{\left|\sqrt{5} x_0+2 y_0\right|}{3} \end{aligned} $

Length of the perpendicular from $P$ to $x-\frac{2}{\sqrt{5}} y=0$ is

$ l_2=\frac{\left|x_0-\frac{2}{\sqrt{5}} y_0\right|}{\sqrt{1+\frac{4}{5}}}=\frac{\left|x_0-\frac{2}{\sqrt{5}} y_0\right|}{\sqrt{\frac{9}{5}}} =\frac{\left|\sqrt{5} x_0-2 y_0\right|}{3} $

$\therefore l_1^2 \cdot l_2^2=\frac{\left[\left(\sqrt{5} x_0+2 y_0\right)^2\left(\sqrt{5} x_0-2 y_0\right)^2\right]}{(3)^2 \times(3)^2}$

$ =\frac{\left(5 x_0-4 y_0\right)^2}{81}=\frac{400}{81} $

$\therefore \quad \frac{l_1^2 l_2^2}{100}=\frac{400}{81 \times 100}=\frac{4}{81}$

The eccentricity of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$ is given by the formula:

$ \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{4}{a^2}} $

The center of this hyperbola is at the origin, $(0, 0)$.

The equation of the circle is:

$ x^2 + y^2 = 16 $

This circle passes through the foci of the hyperbola, which are located at $(ae, 0)$. Thus, we can write:

$ a^2 e^2 = 16 $

Since $e = \sqrt{1 + \frac{4}{a^2}}$, we substitute:

$ a^2\left(1 + \frac{4}{a^2}\right) = 16 $

Expanding and simplifying gives:

$ a^2 + 4 = 16 \quad \Rightarrow \quad a^2 = 12 $

Further simplifying the eccentricity:

$ e = \sqrt{1 + \frac{4}{12}} = \frac{2}{\sqrt{3}} $

Now, let $e'$ be the eccentricity of the conjugate hyperbola. We have:

$ \frac{1}{e^2} + \frac{3}{4} = 1 $

Solving for $e'$:

$ \frac{1}{e'^2} = 1 - \frac{3}{4} \quad \Rightarrow \quad e'^2 = 4 \quad \Rightarrow \quad e' = 2 $

To determine the equation of the tangent line that is tangent to both the hyperbola $x^2 - \frac{y^2}{3} = 1$ and the parabola $y^2 = 8x$, we start with the tangent equation for the parabola. The general equation for a tangent to the parabola $y^2 = 8x$ is:

$ y = mx + \frac{2}{m} $

Since this line is also tangent to the hyperbola, it must satisfy the condition for tangency to the hyperbola as well. The condition for tangency to a hyperbola of the form $x^2 - \frac{y^2}{a^2} = 1$ is given by:

$ c = \pm \sqrt{m^2 - 3} $

Given $c = \frac{2}{m}$, equating this to the tangency condition gives:

$ \frac{2}{m} = \pm \sqrt{m^2 - 3} $

Squaring both sides, we get:

$ \frac{4}{m^2} = m^2 - 3 $

Rearranging terms gives:

$ m^4 - 3m^2 - 4 = 0 $

We can factor this as:

$ (m^2 - 4)(m^2 + 1) = 0 $

Solving for $m^2$, we find:

$ m^2 = 4 \quad \text{(since $m^2 \neq -1$)} $

Thus, $m = \pm 2$. For the tangent with a positive slope:

$ m = 2 $

Substituting $m = 2$ back into the tangent equation of the parabola, we have:

$ y = 2x + \frac{2}{2} = 2x + 1 $

So, the equation of the tangent line with a positive slope is:

$ y - 2x - 1 = 0 $

To find the locus of the midpoints of the chords of the hyperbola $x^2 - y^2 = a^2$ that touch the parabola $y^2 = 4ax$, we begin by considering a point $(h, k)$, which is the midpoint of a chord of the hyperbola.

The equation of this chord in the format $T = S_1$ is:

$ hx - ky = h^2 - k^2 $

Rewriting it in slope-intercept form gives:

$ y = \frac{h}{k}x + \frac{k^2 - h^2}{k} $

Since this line is tangent to the parabola $y^2 = 4ax$, we use the point of tangency conditions:

For a line $y = mx + c$ tangent to the parabola, the condition is:

$ c = \frac{a}{m} $

Substituting $m = \frac{h}{k}$ and $c = \frac{k^2 - h^2}{k}$, we equate and simplify:

$ \frac{k^2 - h^2}{k} = \frac{ak}{h} $

Solving this equation gives:

$ hk^2 - h^3 = ak^2 $

Thus, the required locus of the midpoint $(h, k)$ satisfies:

$ x(y^2 - x^2) = ay^2 $

To find the value of $ b^2 $, we need to consider the properties of the given ellipse and hyperbola. Let's denote $ e_1 $ and $ e_2 $ as the eccentricities of the ellipse and hyperbola, respectively.

For the ellipse given by $\frac{x^2}{16} + \frac{y^2}{b^2} = 1$, the eccentricity $ e_1 $ is calculated as:

$ e_1 = \sqrt{1 - \frac{b^2}{16}} $

For the hyperbola given by $\frac{x^2}{9} - \frac{y^2}{16} = -1$, the eccentricity $ e_2 $ is calculated as:

$ e_2 = \sqrt{1 + \frac{9}{16}} $

We are given that the product of these eccentricities is 1:

$ e_1 \times e_2 = 1 $

Substituting the expressions for $ e_1 $ and $ e_2 $:

$ \sqrt{1 - \frac{b^2}{16}} \times \sqrt{1 + \frac{9}{16}} = 1 $

Squaring both sides to eliminate the square roots, we get:

$ (1 - \frac{b^2}{16})(1 + \frac{9}{16}) = 1 $

Simplifying the equation:

$ \frac{(16 - b^2)(9 + 16)}{16 \times 16} = 1 $

This further simplifies to:

$ 16 - b^2 = \frac{256}{25} $

Solving for $ b^2 $:

$ b^2 = 16 - \frac{256}{25} $

Putting the expression under a common denominator and simplifying gives us:

$ b^2 = \frac{400}{25} - \frac{256}{25} = \frac{144}{25} $

Therefore, $ b^2 = \frac{144}{25} $.

We have, equarion of hyperbola

$ \begin{aligned} & 5 x^2-k y^2=12 \\ & \frac{x^2}{\frac{12}{5}-\frac{y^2}{12}}=1 \end{aligned} $

Here, $a^2=\frac{12}{5}$ and $b^2=\frac{12}{K}$

Equation of tangent of hyperbola is

$ 5 x-2 y-6=0 $

or $ y=\frac{5}{2} x-3 $

Here, $m=\frac{5}{2}$ and $a^2 m^2-b^2=9$

$ \begin{gathered} \frac{12}{5} \times \frac{25}{4}-\frac{12}{K}=9 \Rightarrow 15-9=\frac{12}{K} \\ K=2 \end{gathered} $

Equation of hyperbola is

$ \begin{aligned} & 5 x^2-2 y^2=12 \\ & \frac{x^2}{\frac{12}{5}}-\frac{y^2}{6}=1 \end{aligned} $

Since, $(\sqrt{6}$, p) lie on hyperbola

$ \begin{aligned} & \Rightarrow 5(\sqrt{6})^2-2\left(p^3\right)=12 \\ & \Rightarrow \quad 30-2 p^2=12 \\ & \quad p^2=9 \quad p=23 \\ & \because \quad p<0, \\ & \therefore \quad p=-3 \end{aligned} $

Equation of normal of hyperbola of $(\sqrt{6},-3), 3$

$ \begin{aligned} & \Rightarrow \quad \frac{12 x}{5 \sqrt{6}}-2 y=\frac{12}{5}+6 \\ & \Rightarrow \quad 2 \sqrt{6 x}-10 y=42 \\ & \Rightarrow \quad \sqrt{6} x-5 y=42 \end{aligned} $

The equation of the hyperbola is

$ x^2 - k y^2 = 3 $

which can be rewritten in standard form as:

$ \frac{x^2}{3} - \frac{y^2}{\frac{3}{k}} = 1 $

Here, $ a^2 = 3 $ and $ b^2 = \frac{3}{k} $.

The angle between the asymptotes is given by

$ \tan^{-1} \frac{b}{a} = \frac{\pi}{3} $

This implies:

$ \tan \frac{\pi}{6} = \frac{b}{a} = \frac{\sqrt{\frac{3}{k}}}{\sqrt{3}} = \frac{1}{\sqrt{k}} $

Therefore:

$ \frac{1}{\sqrt{3}} = \frac{1}{\sqrt{k}} \Rightarrow k = 3 $

Recalculate the standard form with $ k = 3 $:

$ 3x^2 - y^2 = 9 $

The pole of the line $ x + y - 1 = 0 $ with respect to the hyperbola $ 3x^2 - y^2 = 9 $ can be found. Let $ (h, k) $ be the pole. The equation of the pole is defined using $ S_1 = 0 $:

$ 3hx - ky = 9 $

This equation is rewritten as:

$ \frac{hx}{3} - \frac{ky}{3} = 1 $

Comparing with the equation $ x + y - 1 = 0 $, we get:

$ \frac{h}{3} = 1 \quad \text{and} \quad \frac{-k}{3} = 1 $

Solving these gives:

$ h = 3 \quad \text{and} \quad k = -9 $

Thus, the pole of the line is $(3, -9)$. The final expression representing the pole in terms of a parameter is:

$ \left(k, \frac{-\sqrt{3}e}{2}\right) $

where $ k = 3 $.

$2 x+\sqrt{6} y=2$ touches $x^2-2 y^2=4$ at

$ \begin{aligned} & P(h, k) \\ & T=0 \Rightarrow h x-2 k y=4 \end{aligned} $

On comparing, $\frac{h}{2}=\frac{-2 k}{\sqrt{6}}=\frac{4}{2}$

$\Rightarrow h=4$ and $k=-\sqrt{6}$

So, $\quad P(h, k)=(4,-\sqrt{6})$

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

∴ Asymptotes of hyperbola

$ y= \pm \frac{b}{a} x $

Then, angle between asymptotes $=30^{\circ}$

$ \begin{aligned} \Rightarrow \tan 30^{\circ} & =\frac{\frac{b}{a}-\left(-\frac{b}{a}\right)}{1+\left(\frac{b}{a}\right)\left(-\frac{b}{a}\right)} \\ & =\frac{(2 b / a)}{\left(1-b^2 / a^2\right)} \\ \Rightarrow \quad \frac{1}{\sqrt{3}} & =\frac{2(b / a)}{1-(b / a)^2} \end{aligned} $

Let $b / a=x$

$ \begin{aligned} &\begin{aligned} & \text { So, } \frac{1}{\sqrt{3}}=\frac{2 x}{1-x^2} \\ & \Rightarrow 1-x^2=2 \sqrt{3} x \\ & \Rightarrow x^2+2 \sqrt{3} x-1=0 \\ & \begin{array}{l} \therefore x=\frac{-2 \sqrt{3} \pm \sqrt{12+4}}{2}=\frac{-2 \sqrt{3} \pm 4}{2} \\ =-\sqrt{3} \pm 2 \\ =2-\sqrt{3} \quad\{\because x>0\} \end{array} \end{aligned}\\ &\text { Therefore, } \frac{b}{a}=2-\sqrt{3}\\ &\begin{aligned} \therefore e & =\sqrt{1+\left(\frac{b}{a}\right)^2}=\sqrt{1+(2-\sqrt{3})^2} \\ & =\sqrt{1+4+3-4 \sqrt{3}}=\sqrt{8-4 \sqrt{3}} \\ & =\sqrt{6}-\sqrt{2} \end{aligned} \end{aligned} $

$ \text { Area of } P A B C=\frac{1}{2}\left|\begin{array}{ll} (x-2) & (-1-1) \\ (y-1) & (-1-2) \end{array}\right| $

$ =\frac{1}{2}|-3(x-2)-2(y-1)| $

$ \begin{aligned} & =\frac{1}{2}|-3 x+6-2 y+2| \\ & =\frac{1}{2}|-3 x-2 y+8| \end{aligned} $

$2 \times$ Area of $\triangle P A B$

$ \begin{aligned} & =2 \times \frac{1}{2}\left|\begin{array}{lll} x & y & 1 \\ 1 & 2 & 1 \\ 2 & 1 & 1 \end{array}\right| \\ & =x(2-1)-y(1-2)+1(1-4) \\ & =x+y-3 \end{aligned} $

According to the question,

$ \begin{gathered} \frac{1}{2}|-3 x-2 y+8|=(x+y-3) \\ \Rightarrow \quad|-3 x-2 y+8|=2(x+y-3) \\ \Rightarrow 9 x^2+4 y^2+64+2(6 x y-16 y-24 x) \\ =4\left(x^2+y^2+9+2(x y-3 y-3 x)\right. \\ \Rightarrow 9 x^2+4 y^2+12 x y-32 y-48 x+64 \\ =4 x^2+4 y^2+36+8 x y-24 y-24 x \\ \Rightarrow 5 x^2+4 x y-8 y-24 x+28=0 \end{gathered} $

$\frac{x^2}{6}-\frac{y^2}{2}=1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For slope of hyperbola

$ \frac{2 x}{6}-\frac{2 y}{2}\left(\frac{d y}{d x}\right)=0 \Rightarrow \frac{d y}{d x}=\frac{x}{3 y} $

Then, slope of normal $=-\frac{3 y}{x}$

From the equation of normal

$ y=-x-n $

Slope $=-1$

So, $\frac{-3 y}{x}=-1 \Rightarrow x=3 y$

By using Eq. (i), we get

$ \begin{array}{rlrl} & & \frac{9 y^2}{6}-\frac{y^2}{2} & =1 \\ \Rightarrow & & y^2 & =1 \Rightarrow y= \pm 1 \\ \text { Then, } & & x & = \pm 3 \\ \text { Now, } & & x y & =-n \\ \Rightarrow & ( \pm 1)( \pm 3 y & =-n \Rightarrow n= \pm 4 \end{array} $

$y=m x+c$ is tangent to hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then point of contact is

$ \left(-\frac{a^2 m}{c},-\frac{b^2}{c}\right) $

Here, $a^2=25, b^2=9$

$ m=m, c=4 $

∴ Point of contact $\left(-\frac{25 \times m}{4},-\frac{9}{4}\right)$ and condition of tangency is

$ \begin{aligned} & \quad c^2=a^2 m^2-b^2 \\ & \Rightarrow \quad 16=25 m^2-9 \Rightarrow m=1(m>0) \\ & \therefore \text { Point of contact }\left(-\frac{25}{4},-\frac{9}{4}\right) \end{aligned} $

The given equation of hyperbola is

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

Now, normal equation at $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ are

$a x \cos \theta+b y \cot \theta=a^2+b^2$

and $a x \cos \phi+b y \cot \phi=a^2+b^2$

Where, $\phi=\frac{\pi}{2}-\theta$ and $(h, k)$ is the point of intersection of the normals drawn at $P$ and $Q$.

So, $a h \cos \theta+b k \cot \theta=a^2+b^2$

and $a h \sin \theta+b k \tan \theta=a^2+b^2$

Now, multiply Eq. (i) by $\sin \theta$ and Eq. (ii) by $\cos \theta$ and subtract them, we get

$ \begin{aligned} & b k(\cos \theta-\sin \theta)=\left(a^2+b^2\right)(\sin \theta-\cos \theta) \\ & \Rightarrow\left(a^2+b^2+b k\right)(\sin \theta-\cos \theta)=0 \\ & \Rightarrow b k+a^2+b^2=0 \quad(\because \sin \theta-\cos \theta \neq 0) \\ & \therefore \quad \quad k=-\frac{\left(a^2+b^2\right)}{b} \quad \forall \theta \end{aligned} $

Hence, the value of $k=-\frac{\left(a^2+b^2\right)}{b}$

To find the equation of the conjugate hyperbola, we are given the equation of a hyperbola as:

$ 9x^2 - 16y^2 + 72x - 32y - 16 = 0 $

First, we'll convert this into the standard form of a hyperbola. Starting with:

$ 9(x^2 + 8x) - 16(y^2 + 2y) = 16 $

Complete the square for both $x$ and $y$:

For $x$:

$ x^2 + 8x \quad \Rightarrow \quad (x+4)^2 - 16 $

For $y$:

$ y^2 + 2y \quad \Rightarrow \quad (y+1)^2 - 1 $

Substituting these back:

$ 9((x+4)^2 - 16) - 16((y+1)^2 - 1) = 16 $

Simplify:

$ 9(x+4)^2 - 144 - 16(y+1)^2 + 16 = 16 $

$ 9(x+4)^2 - 16(y+1)^2 = 144 $

This yields the standard form of the hyperbola:

$ \frac{(x+4)^2}{16} - \frac{(y+1)^2}{9} = 1 $

Now, for the conjugate hyperbola, the equation is:

$ \frac{(x+4)^2}{16} - \frac{(y+1)^2}{9} = -1 $

This can be rewritten as:

$ 9(x+4)^2 - 16(y+1)^2 = -144 $

Expanding this:

$ 9(x^2 + 8x + 16) - 16(y^2 + 2y + 1) = -144 $

Which simplifies to:

$ 9x^2 + 72x + 144 - 16y^2 - 32y - 16 = -144 $

Combining the constants gives:

$ 9x^2 + 72x - 16y^2 - 32y + 288 - 16 = 0 $

Thus, the equation of the conjugate hyperbola is:

$ 9x^2 - 16y^2 + 72x - 32y + 272 = 0 $

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

Focus of parabola : $(ae,\,0)$

Directrix : $x = - ae$.

Equation of parabola $ \equiv {y^2} = 4aex$

Length of latus rectum of parabola $ = 4ae$

Length of latus rectum of hyperbola $ = {{2.{b^2}} \over a}$

as given, $4ae = {{2{b^2}} \over a}\,.\,e$

$2 = {{{b^2}} \over {{a^2}}}$ ...... (i)

$\because$ H passes through $\left( {2\sqrt 2 , - 2\sqrt 2 } \right) \Rightarrow {8 \over {{a^2}}} - {8 \over {{b^2}}} = 1$ ........ (ii)

From (i) and (ii) ${a^2} = 4$ and ${b^2} = 8 \Rightarrow e = \sqrt 3 $

$\Rightarrow$ Equation of parabola is ${y^2} = 8\sqrt 3 x$.