Every normal to $xy = 1$ must have positive slope as $\dfrac {-dx}{dy} = x^{2}$. So $\dfrac {a - 1}{b} > 0$.

$\dfrac {x^2}9-\dfrac {y^2}4=1$.Let $P(x _1, y _1)$ be the point on hyperbola.

The equation of hyperbola is $xy=c^2$ and point $(ct _1,\dfrac{c}{t _1})$ lies on it.

Let us find the equation of the normal.

Equation of hyprbola can be written as $y=\dfrac{c^2}{x}$ and therefore slope of tangent is given by first derivative i.e. $\dfrac{dy}{dx}=−\dfrac{c^2}{x^2}$

hence slope of normal is given by $\dfrac{x^2}{c^2}$ and at $(ct _1,\dfrac{c}{t _1})$ is $t^2$ and its equation is

$y=t _1^2(x−ct _1)+\dfrac{c}{t _1}$

or $xt _1^3−yt _1−ct _1^4+c=0$

As this passes through $(ct _2,\dfrac{c}{t _2})$

$ct _2t _1^3−\dfrac{c}{t _2}t _1−ct _1^4+c=0$

or $ct _1^3(t _2−t _1)+\dfrac{c}{t _2}(t _2−t _1)=0$

as $t _1\neq t _2, t _1−t _2\neq 0$ and dividing by it we get

$ct _2^3=−\dfrac{c}{t _2}$

Or $ t _2^3t _2=−1$

Assume rectangular hyperbola is $xy = c^2$
Thus equation of tangent and normal at any point 't' are,
$\cfrac{x}{t}+ty=2c$ and $ y-\cfrac{c}{t}=t^2(x-ct)$
Now putting $y=0$ in both the equation we get, $x _1=2ct, x _2=ct-\cfrac{c}{t^3}$
and putting $x=0$ we get, $y _1=\cfrac{2c}{t}, y _2=\cfrac{c}{t}-ct^3$
$\Rightarrow x _1x _2+y _1y _2=2ct(ct-\cfrac{c}{t^3})+\cfrac{2c}{t}(\cfrac{c}{t}-ct^3)=0$
Hence, option 'C' is correct.

Given hyperbola is, $\displaystyle \cfrac{x^{2}}{a^2} - \cfrac{y^{2}}{b^2} = 1$
The general equation of normal to hyperbola with slope 'm' is given by,
$y = mx\pm\cfrac{(a^2+b^2)m}{\sqrt{a^2-b^2m^2}}$
Let any external point through wich this line is passing is $P(x _1,y _1)$
$\Rightarrow (y _1-mx _1)^2=\cfrac{(a^2+b^2)^2m^2}{a^2-b^2m^2}$
$\Rightarrow (y _1-mx _1)^2(a^2-b^2m^2)=(a^2+b^2)^2m^2$
Clearly, this is a polynomial of degree four so maximum number of normal that can be drawn from point P (any external point) to the hyperbola is 4 corresponding to four roots of m.
Hence, option 'B' is correct.

The equation of the normal $t _{1}$is  $y-\dfrac{c}{t _{1}}=t _{1}^{2}(x-ct _{1})$

If this passes through $\left ( ct _{2},\dfrac{c}{t _{2}} \right )$

$\dfrac{\mathrm{c}}{\mathrm{t} _{2}}-\dfrac{\mathrm{c}}{\mathrm{t} _{\mathrm{t}}}=\mathrm{t} _{1}^{2}(\mathrm{c}\mathrm{t} _{2}-\mathrm{c}\mathrm{t} _{1})$

$\displaystyle



\Rightarrow-\dfrac{1}{\mathrm{t} _{\mathrm{t}}\mathrm{t} _{2}}=\mathrm{t} _{1}^{2}\Rightarrow

1+\mathrm{t} _{\mathrm{t}}^{3}\mathrm{t} _{2}=0$

$\Rightarrow t _1^3t _2=-1$

Hence, option 'D' is correct.

The equation of the normal at $\left( a\sec { \theta  } ,b\tan { \theta  }  \right) $ to the given hyperbola is
$ax\cos { \theta  } +by\cot { \theta  } =\left( { a }^{ 2 }+{ b }^{ 2 } \right) $
This meets the transverse axis (i.e) at $G$. So, the coordinates of $G$ are $\left{ \cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec { \theta  }
,0 \right} $
The coordinates of the vertices $A$ and $A'$ are $A(a,0)$ and $A'(-a,0)$ respectively
$\therefore

\quad AG.A'G=\left( -a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec {

\theta  }  \right) \left( a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec

{ \theta  }  \right)$
 $\Rightarrow AG.A'G=\left( -a+a{ e }^{ 2 }\sec { \theta  }  \right) \left( a+a{ e }^{ 2 }\sec { \theta  }  \right) $
$={ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta  } -1 \right) $
Hence, option 'B' is correct.

The equation of the normal at $\left( a\sec { \theta  } ,b\tan { \theta  }  \right) $ to the given hyperbola is
$ax\cos { \theta  } +by\cot { \theta  } =\left( { a }^{ 2 }+{ b }^{ 2 } \right) $
This meets the transverse axis (i.e) at $G$. So, the coordinates of $G$ are $\left{ \cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec { \theta  }

,0 \right} $
The coordinates of the vertices $A$ and $A'$ are $A(a,0)$ and $A'(-a,0)$ respectively
$\therefore

\quad AG.A'G=\left( -a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec {

\theta  }  \right) \left( a+\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ a } \sec

{ \theta  }  \right)$
 $\Rightarrow AG.A'G=\left( -a+a{ e }^{ 2 }\sec { \theta  }  \right) \left( a+a{ e }^{ 2 }\sec { \theta  }  \right) $
$={ a }^{ 2 }\left( { e }^{ 4 }\sec ^{ 2 }{ \theta  } -1 \right) $
Hence, option 'B' is correct.

The equation of the normal at $P\left( a\sec { \theta  } ,b\tan {

\theta  }  \right) $ to the hyperbola $\cfrac { { x }^{ 2 } }{ { a

}^{2 } } -\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ is
$ax\cos { \theta  } +by\cot { \theta  } ={ a }^{ 2 }+{ b }^{ 2 }$
This intersects the transverse and conjugate axes at $ L\left( \cfrac { {

a}^{ 2 }+{ b }^{ 2 } }{ a } \sec { \theta  } ,0 \right) $ and

$M\left(0,\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{  } } \tan {

\theta  }  \right) $ respectively
Let $N(h,k)$. then $h=\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ b } \sec { \theta  } $ and
$k=\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ { b }^{  } } \tan { \theta  } $
$\Rightarrow

\sec { \theta  } =\cfrac { 2ah }{ { a }^{ 2 }+{ b }^{ 2 } } \quad

,\quad \tan { \theta  } =\cfrac { 2bk }{ { a }^{ 2 }+{ b }^{ 2 } }

\quad$
$\therefore \sec ^{ 2 }{ \theta  } -\tan ^{ 2 }{ \theta  }

=1\quad \Rightarrow 4{ a }^{ 2 }{ h }^{ 2 }-4{ b }^{ 2 }{ k }^{ 2 }={

\left( { a }^{ 2 }+{ b }^{ 2 } \right)  }^{ 2 }$
Thus the locus of

$(h,k)$ is $\Rightarrow 4{ a }^{ 2 }{ x }^{ 2 }-4{ b }^{ 2 }{ y }^{

2}={ \left( { a }^{ 2 }+{ b }^{ 2 } \right)  }^{ 2 }\quad $
Let ${ e } _{ 1 }$ be the eccentricity of this hyperbola. Then
${{

e } _{ 1 } }^{ 2 }=1+\cfrac { { a }^{ 2 } }{ { b }^{ 2 } } =\cfrac { {a

}^{ 2 }+{ b }^{ 2 } }{ { b }^{ 2 } } =\cfrac { { a }^{ 2 }{ e }^{ 2 }}{ {

a }^{ 2 }({ e }^{ 2 }-1) } $
$\Rightarrow { e } _{ 1 }=\displaystyle \frac { e }{ \sqrt { { e }^{ 2 }-1 }  } $

Given hyperbola is, $\displaystyle \cfrac{x^{2}}{a^2} - \cfrac{y^{2}}{b^2} = 1$
The general equation of normal to hyperbola with slope $m$ is given by,
$y = mx\pm\cfrac{(a^2+b^2)m}{\sqrt{a^2-b^2m^2}}$
Let any external point through wich this line is passing is $P(x _1,y _1)$
$\Rightarrow (y _1-mx _1)^2=\cfrac{(a^2+b^2)^2m^2}{a^2-b^2m^2}$
$\Rightarrow (y _1-mx _1)^2(a^2-b^2m^2)=(a^2+b^2)^2m^2$
Clearly, this is a polynomial of degree four so maximum number of normal that can be drawn from point $P$ (any external point) to the hyperbola is $4$ corresponding to four roots of $m$.
Hence, option 'B' is correct.

Equations of the normal at P is $ax+bycosec\theta =\left ( a^{2}+b^{2} \right )\sec \theta $          (i)
and the equation of the normal at $Q\left ( a\sec \phi , b\sec \phi  \right )$ is
$ax+by cosec\phi =\left ( a^{2}+b^{2} \right )\sec \phi $          (ii)
Subtracting (ii) from (i) we get

   $\displaystyle y=\frac{a^{2}+b^{2}}{b}.\frac{\sec \theta -\sec \phi }{cosec \theta -cosec \phi }$

So $\displaystyle k=y=\frac{a^{2}+b^{2}}{b}.\frac{\sec \theta -\sec

\left ( \pi /2-\theta  \right )}{cosec \theta -cosec \left ( \pi

/2-\theta  \right )}$          $\left [ \because \theta +\phi =\pi /2

\right ]$

     $\displaystyle =\frac{a^{2}+b^{2}}{b}.\frac{\sec

\theta -cosec \theta }{cosec \theta -\sec \theta }=-\left [

\frac{a^{2}+b^{2}}{b} \right ]$

We know equation of normal to the hyperbola $\cfrac{x^2}{a^2}-\cfrac{y^2}{b^2}=1$ is given by, $\cfrac{a^2x}{x _1}+\cfrac{b^2y}{y _1}=a^2-b^2$

Thus, required normal is, $5x+\cfrac{16y}{0}=9$
Clearly denominator of second term of L.H.S is $0$ so the equation of line is, $y=0$ 
Hence, option 'A' is correct.  

Required normal is given by,
$\displaystyle \frac{a^2x}{x _1}+\frac{b^2y}{y _1}=a^2+b^2$
$\Rightarrow \displaystyle \frac{16x}{6}+\frac{9y}{(3\sqrt{5}/2)}=25$
$\Rightarrow 8\, \sqrt{5}x\, +\, 18y\, =\, 75\, \sqrt{5}$

Suppose $\displaystyle \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ be a hyperbola and let $\displaystyle \dfrac{x^2}{a^2} - \frac{y^2}{b^2} = - 1$ be its conjugate.
Then their eccentricities are given by $e^2 = \displaystyle \dfrac{a^2 + b^2}{a^2}$ and $\displaystyle e'^2 = \frac{a^2 + b^2}{b^2}$ respectively.
$\therefore \displaystyle \dfrac{1}{e^2} + \dfrac{1}{e'^2} = \dfrac{a^2}{a^2 + b^2} + \dfrac{b^2}{a^2 + b^2} = 1$

Equation of normal $\displaystyle Y-y=-\frac { dy }{ dx } \left( X-x \right) $
$\displaystyle \Rightarrow G=\left( x+y\frac { dy }{ dx } ,0 \right) $
According to question
$\displaystyle \left| x+y\frac { dy }{ dx }  \right| =\left| 2x \right| \Rightarrow y\frac { dy }{ dx } =x$ or $\displaystyle y\frac { dy }{ dx } =-3x$
$\Rightarrow ydy=xdx$ or $ydy=-3xdx$
$\displaystyle \Rightarrow \frac { { y }^{ 2 } }{ 2 } =\frac { { x }^{ 2 } }{ 2 } +c$ or $\displaystyle \frac { { y }^{ 2 } }{ 2 } =-\frac { 3{ x }^{ 2 } }{ 2 } +c$
$\Rightarrow { x }^{ 2 }-{ y }^{ 2 }=-2c$ or $3{ x }^{ 2 }+{ y }^{ 2 }=2c$

$xt^{3}-yt-t^{4}+120$
$eq^{n}$ of the normal at $t$ will same as the line $ax+by+c=0$
$\therefore \dfrac{t^{3}}{a}=\dfrac{-t}{b}=\dfrac{-1-t^{4}}{c}$
$t^{2}=\dfrac{-a}{b}$
$\therefore ab<0$

$\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$   $\Rightarrow \dfrac{2x}{16} - \dfrac{2y}{9} \dfrac{dy}{dx} = 0$

equation of hyperbola at point $A(2\sec{\theta} , 3\tan{\theta})$ is

The equations of the normal at P is $ax+bycosec\theta =\left ( a^{2}+b^{2} \right )\sec \theta $          (i)

and the equation of the normal at $Q\left ( a\sec \phi , b\sec \phi  \right )$ is
$ax+by cosec\phi =\left ( a^{2}+b^{2} \right )\sec \phi $          (ii)
Subtracting (ii) from (i) we get

   $\displaystyle y=\frac{a^{2}+b^{2}}{b}.\frac{\sec \theta -\sec \phi }{cosec \theta -cosec \phi }$

So $\displaystyle k=y=\frac{a^{2}+b^{2}}{b}.\frac{\sec \theta -\sec

\left ( \pi /2-\theta  \right )}{cosec \theta -cosec \left ( \pi

/2-\theta  \right )}$          $\left [ \because \theta +\phi =\pi /2

\right ]$
    
$\displaystyle =\frac{a^{2}+b^{2}}{b}.\frac{\sec

\theta -cosec \theta }{cosec \theta -\sec \theta }=-\left [

\frac{a^{2}+b^{2}}{b} \right ]$
Hence, option 'D' is correct.

Equation of normal with slope $'m'$ to the parabola $y^2=4ax$ is given by,
$y=mx-2am-am^3$
Also this line touches the hyperbola $x^2-y^2=a^2$
thus using condition of tangency to the hyperbola, $c^2=a^2m^2-b^2$
$(-2am-am^3)^2=a^2(m^2-1)$
$\Rightarrow 4m^2+m^6+4m^4=m^2-1$
$\Rightarrow m^6+4m^4+3m^2+1=0$
Hence, option 'D' is correct.

Equation of normal with slope $'m'$ to the parabola $y^2=4ax$ is given by,
$y = mx-2am-am^3$ (i)

Normal at $\theta,\, \phi$ are
$\displaystyle \left {

\begin{matrix} ax\, cos\, \theta & + & by\, cot\, \theta\, =\,

a^2\, +\, b^2 \ ax\, cos\, \phi & + & by\, cot\, \phi\, =\,

a^2\, +\, b^2 \end{matrix}\right.$
where $\displaystyle \phi\, =\, \frac{\pi}{2}\, -\, \theta$ and these passes through (h, k).

$\therefore\, ah\, cos \theta\, +\, bk\, cot \theta\, =\, a^2\, +\, b^2$ .....(i)
$ah\, sin \theta\, +\, bk\, tan \theta\, =\, a^2\, +\, b^2$ .....(ii)
Multiply (i) by $sin \theta$ & (ii) by $cos \theta$ & subtract them, 
we get
$\Rightarrow\, (bk\, +\, a^2\, +\, b^2)\, (sin \theta\, -\, cos \theta)\, =\, 0$
$k =-(\cfrac{a^2 + b^2}{b})$
Hence, option 'D' is correct.

Clearly tangent at P and Q intersect at right-angles at S (say)
$ \displaystyle \Rightarrow $ PSQR is cyclic