The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,

Equation of hyperbola is $\dfrac { { x }^{ 2 } }{ 64 } -\dfrac { { y }^{ 2 } }{ 49 } =1$

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,

Solving given equation we get,
$\displaystyle \tan\theta=\frac{ax-by}{ay+bx}$
and $\displaystyle \sec\theta = \frac{x^2+y^2}{ay+bx}$
Eliminating $\theta$ we get, $x^2+y^2=a^2+b^2$ which is director  circle of the ellipse $\displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2} = 1$

The director circle is the locus of the point of intersection of a pair of perpendicular tangents to a hyperbola.

Equation of the director circle of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2  b^2$ i.e. a circle whose center is origin and radius is $(a^2  b^2)$.

Hence, for the director circle to be imaginary, $a^2 < b^2$ .

$-p^2 + 6p + 5 < -p - 3$ ---> $p^2 -7p -8 > 0$ - $(p-  8)(p + 1) > 0$ $\Rightarrow$ $p$ lies in $R - [-1 , 8]$. Hence, option (A).

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$,

Let $P$ be $(h,k)$
Now Chord of contact of tangent from $P$ to the hyperbola $x^2-y^2=a^2$ is,
$T =0\Rightarrow hx -ky = a^2$ (i)
And director circle of given hyperbola is, $x^2+y^2=a^2$
Thus equation of chord of contact to this circle from P is, $hx+ky = a^2$ (ii)
Now given line (i) and (ii) are perpendicular,
$\Rightarrow \cfrac{h}{k}\times \cfrac{-h}{k}=-1\Rightarrow h^2=k^2$
Hence locus of $P$ is given by, $x^2-y^2=0$ 

Since, the circle $x^2\, +\, y^2\, =\, a^2$ intersects the hyperbola $xy\, =\, c^2$
Therefore, $x^2+\dfrac{c^4}{x^2}=a^2$
$\Rightarrow x^4-a^2x^2+c^4=0$
now sum of the roots: $x _1+x _2+x _3+x _4=0$
and product of the roots $x _1x _2x _3x _4=c^4$
Repeat the same for $y$

Given hyperbola is $\dfrac{x^2}{9}-\dfrac{y^2}{b}=1$

for hyperbola,

for hyperbola, $\dfrac{x^2}{36}- \dfrac{y^2}{16}=1$ 

The Auxiliary circle of a hyperbola is defined as the circle with the center same as the hyperbola and with transverse axis as it's diameter. The end points of transverse axis are the two vertices of the hyperbola, so the circle also contains the two vertices of the hyperbola.

For any Hyperbola of the form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

The equation of director circle whose center us at origin and radius $\sqrt{a^2-b^2}$

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

for hyperbola $a > b$

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

For a standard hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, the equation of director circle is given by:

$\begin{array}{l} equation\, \, of\, \, director\, \, circle\, \, { x^{ 2 } }+{ y^{ 2 } }={ a^{ 2 } }+4ab-\left( { 2ab-{ b^{ 2 } } } \right)  \ \Rightarrow { x^{ 2 } }+{ y^{ 2 } }={ a^{ 2 } }+{ b^{ 2 } }+2ab \ \Rightarrow { x^{ 2 } }+{ y^{ 2 } }={ \left( { a+b } \right) ^{ 2 } } \ Then,\, \, Radians\, \, of\, \, circle\, \, is\left( { a+b } \right)  \end{array}$

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

The Director circle of a hyperbola is defined as the locus of the point of intersection of two perpendicular tangents to the hyperbola. For any standard hyperbola $\dfrac{x^2}{a^2} -\dfrac {y^2}{b^2} = 1$,

$P(a sec(t) , a tan(t))$ and $P'(a sec(t) , 0)$
$|PP'| = a^2 tan^2(t)$
The power of point P' relative to a circle $x^2 + y^2 = a^2$ is :

$(asec(t) - 0)^2 + (0-0)^2 - a^2$   (power of the point w.r.t. circle)

$= a^2sec^2(t) - a^2 = a^2(sec^2(t)-1) = a^2tan^2(t)$

Let $P\left( { x } _{ 1 },{ y } _{ 1 } \right) $ be the pole of the line

$lx+my+n=0$ with respect ot the hyperbola $\cfrac { { x }^{ 2 } }{ {

a }^{ 2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$
Then the equation of the polar is
$\cfrac

{ { xx } _{ 1 } }{ { a }^{ 2 } } -\cfrac { { yy } _{ 1 } }{ { b }^{ 2 } }

=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$
Since $\left( { x } _{ 1 },{ y } _{ 1 } \right) $  is the pole of the line
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$
Clearly $(i)$ and $(ii)$ represent the same line. Therefore,
$\therefore \quad \cfrac { { x } _{ 1 } }{ { a }^{ 2 }l } =\cfrac { { -y } _{ 1 } }{ { b }^{ 2 }m } =\cfrac { 1 }{ -n } $
${ x } _{ 1 }=\cfrac { { -a }^{ 2 }l }{ n } ,\quad { y } _{ 1 }=\cfrac { { b }^{ 2 }m }{ n } $
Hence the pole of the given line with respect of the given hyperbola is
$\left(- \cfrac { { a }^{ 2 }l }{ n } ,\cfrac { { b }^{ 2 }m }{ n }  \right) \quad $

The tangent equation to the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ is 
$y=mx \pm \sqrt{a^2m^2-b^2}$
$\Rightarrow (y-mx)=\pm \sqrt{a^2m^2-b^2}$
On squaring both sides, we get
$\Rightarrow (y-mx)^2=(a^2m^2-b^2)$
$\Rightarrow (x^2-a^2)m^2-2xym+(y^2+b^2)=0$
Product of the slopes,$m _1m _2=\dfrac{(y^2+b^2)}{x^2-a^2}$
But given tangents are perpendicular to each other $\Rightarrow$ Their product of slopes equal to $-1.$
$\Rightarrow m _1m _2=-1$
$\Rightarrow \dfrac{(y^2+b^2)}{x^2-a^2}=-1$
$\Rightarrow x^2+y^2=(a^2-b^2)$
But given hyperbola equation as $\dfrac{x^2}{\cos\alpha^2}-\dfrac{y^2}{\sin\alpha^2}=1$
The required tangent equation is $x^2+y^2=\cos^2\alpha-\sin^2\alpha=\cos 2\alpha$
Since radius of circle is always greater than equal to zero.
$\Rightarrow \cos 2\alpha \geq 0$
But maximum value of $\cos$ is $1$.
$\Rightarrow 0 \leq \cos2\alpha \leq 1$
$\Rightarrow  \dfrac{\pi}{2} \leq 2\alpha \leq 0$
$\Rightarrow  \dfrac{\pi}{4} \leq \alpha \leq 0$
$\Rightarrow \alpha$has inifinite number of solutions.

Equation of any tangent in terms of slope $m$ is

$y = mx + (a^2m^2  b^2)$

It passes through $(h, k)$, so we have

$(k - mh)^2 = a^2m^2 - b^2$

So, $m^2(h^2 - a^2) - 2mhk + k^2 + b^2 = 0$

This is a quadratic in $m$

Let the slopes of tangents be $m _1$ and $m _2$.

then $m _1.m _2 = -1$.

So, $\dfrac{(k^2 + b^2)}{(h^2 - a^2)} = -1$

$(h^2 + k^2) = (a^2  b^2)$

Hence, the locus is $(x^2 + y^2) = (a^2  b^2)$ which is the director circle of $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1.$

Equation of tangent of given hyperbola at point
$\displaystyle (h,\, k)\,$ is $\dfrac{hx}{a^2}\, -\, \dfrac{ky}{b^2}\, =\, 1$ ...(i)
Equation of auxillary circle is $x^2\, +\, y^2\, =\, a^2$ .....(ii)
From (i) and (ii)
$\displaystyle \left [ \left ( 1\, +\, \frac{ky}{b^2}\right ) \frac{a^2}{h}\right ]^2\, +\, y^2\, -\, a^2\, =\, 0$
$\Rightarrow\, y^2\, (k^2a^4\, +\, b^4h^2)\, +\, 2kb^2a^4y\, +\, b^4a^2\, (a^2\, -\, h^2)\, =\, 0$
Now  $\displaystyle \, \frac{y _1\, +\, y _2}{y _1y _2}\, =\, -\,

\frac{2kb^2a^4}{b^4a^2(a^2\, -\, h^2)}\, =\, \frac{-2ka^2}{b^2a^2 \left (

1\, -\, \frac{h^2}{a^2}\right )}$
$\displaystyle =\, \frac{-2k}{b^2 \left ( \frac{-k^2}{b^2}\right )}\, =\, \frac{2}{k}$

The director circle is the locus of the point of intersection of a pair of perpendicular tangents to a hyperbola.

Equation of the director circle of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ is $x^2 + y^2 = a^2  b^2$ i.e. a circle whose center is origin and radius is $(a^2  b^2)$.

Hence, for the director circle to be a point circle, $a^2 = b^2$ .

$p^2 - 4 = p^2 + 4p + 3$ ---> $4p = -7$ ---> $p = -\dfrac{7}{4}$ . Hence, option (B).