Answer:
Option D
Explanation:
We have,
$\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ and $\frac{x^{2}}{16}-\frac{y^{2}}{k}=1$
On solving these equation , we get
$x^{2}=\frac{144+16k}{36+k}$ and $y^{2}= \frac{-27k}{36+k}$ ..........(i)
Now, $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$
$\Rightarrow$ $\frac{2x}{4}+\frac{2yy'}{9}=0\Rightarrow y'=-\frac{9}{4}\frac{x}{y}$ ..........(ii)
again , $\frac{x^{2}}{16}-\frac{y^{2}}{k}=1 \Rightarrow \frac{2x}{16}-\frac{2yy'}{k}=0$
$\Rightarrow$ $y'= \frac{k}{16} \frac{x}{y}$ ............(iii)
Since both curves are orthogonal
$\therefore$ $\frac{-9}{4}\frac{x}{y}\times\frac{k}{16}\frac{x}{y}=-1\Rightarrow 9kx^{2}=64y^{2}$
From Eq.(i) , we have
$9k\left(\frac{144+16k}{36+k}\right)=64\left(\frac{-27k}{36+k}\right)\Rightarrow k=-21$
