We have,
$$
\frac{x^2}{4}+\frac{y^2}{9}=1 \text { and } \frac{x^2}{16}-\frac{y^2}{k}=1
$$
On solving these equation, we get
$$
x^2=\frac{144+16 k}{36+k} \text { and } y^2=\frac{-27 k}{36+k}
$$
Now, $\quad \frac{x^2}{4}+\frac{y^2}{9}=1$
$\Rightarrow \quad \frac{2 x}{4}+\frac{2 y y^{\prime}}{9}=0 \Rightarrow y^{\prime}=-\frac{9}{4} \frac{x}{y}$
Again, $\frac{x^2}{16}-\frac{y^2}{k}=1 \Rightarrow \frac{2 x}{16}-\frac{2 y y^{\prime}}{k}=0$
$\Rightarrow \quad y^{\prime}=\frac{k}{16} \frac{x}{y}$
Since both curves are orthogonal
$$
\therefore \quad \frac{-9}{4} \frac{x}{y} \times \frac{k}{16} \frac{x}{y}=-1 \Rightarrow 9 k x^2=64 y^2
$$
From Eq. (i), we have
$$
9 k\left(\frac{144+16 k}{36+k}\right)=64\left(\frac{-27 k}{36+k}\right) \Rightarrow k=-21
$$