Match the item of column I with those of column II.

Column I		Column II
(i)	The locus of the point whose chord of contact with respect to the hyperbole $\frac{x^2}{16}-\frac{y^2}{9}=1$ touches the circle describe on the line joining the foci is	(A)	${\left(x^2+y^2\right)}^2=4x^2-3y^2$
(ii)	The chords of the circle  $x^2+y^2=4$ touches the hyperbola $\frac{x^2}{4}-\frac{y^2}{3}=1$ Then, the locus of the mid-point of these chords is	(B)	$x^2+y^2=9$
(iii)	The director circle of the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ is	(C)	$x^2-y^2=32$
(iv)	The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$. Then, the equations of the hyperbola is	(D)	$\frac{x^2}{{16}^2}+\frac{y^2}{9^2}=\frac{1}{25}$
		(E)	$x^2+y^2=9^2$

i) $P\left(x_1,y_1\right)$ Be a point whose chord of contact

$\frac{x{x}_1}{16}-\frac{y{y}_1}{9}=1$

touches the circle described on the line joining the foci  $S\left(-5,0\right)$ and $S\text{'}\left(-5,0\right)$ whose equal is $x^2+y^2=25$

$\Rightarrow \left|\frac{0-0-1}{\sqrt{x_1^2/{16}^2+y_1^2/9^2}}\right|=5$

$\Rightarrow \frac{x_1^2}{{16}^2}+\frac{y_1^2}{9^2}=\frac{1}{25}$

Locus is  $\frac{x^2}{{16}^2}+\frac{y^2}{9^2}=\frac{1}{25}$

ii) $\left(x_1,y_1\right)$ is mid-point of a chord  $x^2+y^2=4$ touching the hyperbola $\frac{x^2}{4}-\frac{y^2}{3}=1$

That is line  $x{x}_1+y{y}_1=x_1^2+y_1^2$ touches the hyperbola

$\Rightarrow {\left(\frac{x_1^2+y_1^2}{y_1}\right)}^2=4{\left(\frac{-x_1}{y_1}\right)}^2=-3$

$\Rightarrow {\left(x_1^2+y_1^2\right)}^2=4x_1^2-3y_1^2$

$\therefore$ Locus is  ${\left(x^2+y^2\right)}^2=4x^2-3y^2$

(iii)Director circle of  $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

$x^2+y^2=a^2-b^2$

Hence, $a^2=25$ and  $b^2=16$

(iv)  Since,  $\sqrt{2}$ is the eccentricity of the hyperbola, it must be a rectangular hyperbola.

$\therefore$ It is of the form $x^2-y^2=a^2$ by hypothesis

$2ae=16$

$\Rightarrow 2a\left(\sqrt{2}\right)=16$

$\Rightarrow a=4\sqrt{2}$

$\therefore$ Hyperbola is  $x^2-y^2=32$