Column-I		Column-II
A)	If the straight line $x+y=\sqrt{2}p,$ touch the hyperbola $4x^2-9y^2=36,$then $2p^2$  is	p)	$2{\tan }^{-1}\frac{3}{4}$
B)	The angle between the two asymptotes of the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ is:	q)	5
C)	The length of the latus rectum of the hyperbola $xy-3x-3y+7=0$ is $l$  then  $2l$ is	r)	8
D)	The number of normals to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$  from an external point is	s)	4

A) Given equation of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$  
Here, $a^2=9,b^2=4$ 
And line  $y=-x+\sqrt{2}P\mathrm{....................}\left(i\right)$
If the line $y = mx+ c$ will touch the hyperbola
$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then  $c^2=a^2{m}^2-b^2\dots .\left(ii\right)$
From the eq. (i), we get
 $m=-1,c=\sqrt{2}p$ 
Putting these values in eq.(ii)
${\left(\sqrt{2}P\right)}^2=9\left(1\right)-4\Rightarrow 2P^2=5$ 
B) We know that angle between two asymptotes of the hyperbola  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is  $2{\tan }^{-1}\left(\frac{b}{a}\right)$
Equation to the hyperbola  $\frac{x^2}{16}-\frac{y^2}{9}=1$
Here, $a = 4, b = 3$
 Required angle =  $2{\tan }^{-1}\left(\frac{3}{4}\right)$
C) Given equation of hyperbola can be rewritten a  $x(y-3)-3(y-3)=2\Rightarrow (x-3)(y-3)=2$
Let $x–3=X and y–3=Y$
Equation of hyperbola is of the form $XY=2$ (rectangular hyperbola). In rectangular hyperbola $a= b,$ so length of latus rectum  $=\frac{2b^2}{a}=2a$ (distance between vertices)
And  $xy=c^2\Rightarrow 2=\frac{a^2}{2}\Rightarrow a=2$
$\therefore$  Length of latus rectum is $2a = 4$