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 $.

Let $p(h, k)$ be the point of intersection of tangents at the ends of a normal chord of the hyperbola, $x^2-y^2=a^2$, then the equation of the chord is

$h x-k y=a^2 \quad \text{... (i)}$

But it is normal chord. So, its equation must be of the form $x\cos\theta+y\cot\theta=2a\quad \text{... (ii)}$

Eqs. (i) and (ii) represents the same line

$\begin{array}{ll} \therefore & \frac{\cos \theta}{h}=\frac{\cot \theta}{-k}=\frac{2 a}{a^2} \\ & \sec \theta=\frac{a}{2 h^{\prime}} \cdot \tan \theta=\frac{-a}{2 k} \\ \therefore & \sec ^2 \theta-\tan ^2 \theta=1 \\ \Rightarrow & \frac{a^2}{4 h^2}-\frac{b^2}{4 k^2}=1 \\ \Rightarrow & a^2\left(k^2-h^2\right)=4 h^2 k^2 \end{array}$

Therefore, the locus of $p(h, k)$ is $a^2\left(y^2-x^2\right)=4 x^2 y^2$

Given, hyperbola is $16 x^2-9 y^2=1$

$\begin{array}{ll} & \frac{x^2}{\frac{1}{16}}-\frac{y^2}{\frac{1}{9}}=1 \\ \Rightarrow \quad & \frac{x^2}{\left(\frac{1}{4}\right)^2}-\frac{y^2}{\left(\frac{1}{3}\right)^2}=1 \end{array}$

We know that Eccentricity of a hyperbola, $e_1^2=1+\frac{b^2}{a^2}$

$\begin{aligned} & \Rightarrow e_1^2=1+\frac{1 \times 16}{9 \times 1} \\ & \Rightarrow e_1^2=\frac{25}{9} \\ & \Rightarrow e_1=\frac{5}{3} \quad \text{.... (i)} \end{aligned}$

Now, eccentricity of its conjugate

$ \begin{array}{rlrl} & e_2^2 =1+\frac{a^2}{b^2} \\ \Rightarrow & e_2^2 =1+\frac{1 \times 9}{16 \times 1} \\ \Rightarrow & e_2^2 =\frac{25}{16} \\ \Rightarrow & e_2=\frac{5}{4} \end{array}$

From Eqs. (i) and (ii),

$\begin{aligned} 3 e_1 & =3 \times \frac{5}{3}=\frac{5}{4} \times 4 \\ \Rightarrow \quad 3 e_1 & =4 e_2 \quad \left[\because e_2=\frac{5}{4}\right] \end{aligned}$

Let $P(\sec \theta, \tan \theta)$ be the point where the normal to the rectangular hyperbola $x^2-y^2=1$ Meet equation of normal at $\frac{\pi}{4}$ is

$\begin{aligned} & \frac{x}{\sec \frac{\pi}{4}}+\frac{y}{\tan \frac{\pi}{4}}=2 \\ \Rightarrow & \frac{x}{\sqrt{2}}+y=2 \\ \Rightarrow \quad & x+\sqrt{2} y=2 \sqrt{2} \end{aligned}$

Substitute $x=2 \sqrt{2}-\sqrt{2} y$ in hyperbola's equation,

$\begin{aligned} & {[\sqrt{2}(2-y)]^2-y^2=1 } \\ \Rightarrow & 2\left[4+y^2-4 y\right]-y^2=1 \\ \Rightarrow & 8+2 y^2-8 y-y^2-1=0 \\ \Rightarrow & y^2-8 y+7=0 \\ \Rightarrow & y=7,1 \\ & x+7 \sqrt{2}=2 \sqrt{2} \text { or } x+\sqrt{2}=2 \sqrt{2} \\ \Rightarrow & x=-5 \sqrt{2} \text { or } x=\sqrt{2} \end{aligned}$

The two points are $(-5 \sqrt{2}, 7),(\sqrt{2}, 1)$.

At $\left(\frac{\pi}{4}\right)$, the point is $(\sqrt{2}, 1)$ and at $\theta$, the point is $(-5 \sqrt{2}, 7)$.

$\begin{aligned} & \sec \theta=-5 \sqrt{2}, \tan \theta=7 \\ & \sec ^2 \theta+\tan \theta=(-5 \sqrt{2})^2+7=57 \end{aligned}$

Vertices and foci of the hyperola are $( \pm 3,0)$ and $( \pm 4,0)$.

Thus, $a= \pm 3$ and $a e= \pm 4$ and $e=\frac{4}{3}$

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

$\frac{16}{9}=1+\frac{b^2}{9} \Rightarrow b^2=7$

Thus, equation of hyperbola is $\frac{x^2}{9}-\frac{y^2}{7}=1$

whose parametric equation is $x=3 \sec \theta$ and $y=\sqrt{7} \tan \theta$

$\begin{aligned} & \frac{1+\tanh x}{1-\tanh x} \\ & =\frac{1+\frac{e^x-e^{-x}}{e^x+e^{-x}}}{1-\frac{e^x-e^{-x}}{e^x+e^{-x}}} \quad\left[\because \tanh x=\frac{e^x-e^{-x}}{e^x+e^{-x}}\right] \\ & =\frac{e^x+e^{-x}+e^x-e^{-x}}{e^x+e^{-x}-e^x+e^{-x}} \\ & =\frac{2 e^x}{2 e^{-x}}=e^{2 x} \end{aligned}$

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

Centre : $O(0,0)$

$\begin{aligned} & \text { Foci : } S( \pm a e, o) \equiv( \pm 3,0) \\ & \Rightarrow \quad a e=3 \\ \end{aligned}$

Eccentricity, $e=\sqrt{1+\left(\frac{b}{a}\right)^2}=\frac{3}{2}$

$\begin{aligned} & \because \quad a e=3 \\ & \Rightarrow \quad a .\left(\frac{3}{2}\right)=3 \\ & \Rightarrow \quad a=2 \\ & \text { and } \sqrt{1+\left(\frac{b}{a}\right)^2}=\frac{3}{2} \\ \end{aligned}$

$\begin{aligned} & \Rightarrow 1+\left(\frac{b}{2}\right)^2=\frac{9}{4} \\ & \Rightarrow \quad \frac{b^2}{4}=\frac{5}{4} \\ & \Rightarrow \quad b^2=5 \\ \end{aligned}$

$\therefore$ Hyperbola equation is $\frac{x^2}{4}-\frac{y^2}{5}=1\quad \text{.... (i)}$

Given, line is $2 x-y=1$

Put $y=2 x-1$ into Eq. (i),

$\begin{aligned} & \frac{x^2}{4}-\frac{(2 x-1)^2}{5}=1 \\ & \Rightarrow 5 x^2-4\left(4 x^2+1-4 x\right)=20 \\ & \Rightarrow 5 x^2-16 x^2-4+16 x-20=0 \\ & \Rightarrow-11 x^2+16 x-24=0 \end{aligned}$

Here, $D=b^2-4 a c=(16)^2-4(-11)(-24)<0$

$\therefore$ Line does not intersect the hyperbola.

Equation of hyperbola

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

Let $(h, k)$ be the point whose chord of contact subtends a right angle at the centre of hyperbola. Equation of chord of contact

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

Squaring both the sides, we get

$\begin{aligned} & \left(\frac{h x}{a^2}\right)^2+\left(\frac{k y}{b^2}\right)^2-2 \frac{h x}{a^2} \cdot \frac{k y}{b^2}=(l)^2 \\ \Rightarrow \quad & \frac{h^2 x^2}{a^4}+\frac{k^2 y^2}{b^4}-\frac{2 k h}{a^2 b^2} \cdot x y=1 \quad \text{... (ii)} \end{aligned}$

Now, from Eqs. (i) and (ii), we have

$\begin{aligned} & \frac{h^2 x^2}{a^4}+\frac{k^2 y^2}{b^4}-\frac{2 h k}{a^2 b^2} \cdot x y=\frac{x^2}{a^2}-\frac{y^2}{b^2} \\ \Rightarrow \quad & \left(\frac{h^2}{a^4}-\frac{1}{a^2}\right) x^2+\left(\frac{k^2}{b^4}+\frac{1}{b^2}\right) y^2-\frac{2 h k}{a^2 b^2} \cdot(x y)=0 \end{aligned}$

The above equation represents a pair of perpendicular lines if coefficient of $x^2$ + coefficient of $y^2=0$

$\begin{aligned} & \Rightarrow \quad \frac{h^2}{a^4}-\frac{1}{a^2}+\frac{k^2}{b^4}+\frac{1}{b^2}=0 \\ & \Rightarrow \quad \frac{h^2}{a^4}+\frac{k^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2} \\ & \text { Replacing } h \rightarrow x \text { and } k \rightarrow y, \\ & \frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2} \end{aligned}$

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

Image

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 $=a b$

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