If the curves $\frac{{\mathrm{x}}^2}{k}+\frac{ {\mathrm{y}}^2}{4}=1$ and ${\mathrm{y}}^3=16\mathrm{x}$ intersect at right angles, then a value of $k$ is

$$\begin{gathered} \frac{x^2}{k}+\frac{y^2}{4}=1\_\_\_\_\left(1\right) \\ {y}^3=16x\_\_\_\left(2\right);P(x,y) \\ Diff. Equ \left(1\right) \Rightarrow \frac{2x}{k}+\frac{y}{2}\frac{dy}{dx}=0 \\ \Rightarrow \frac{dy}{dx}=\frac{-4x}{ky}=m_1 \end{gathered}$$
$$\begin{gathered} Diff. Equ \left(2\right)\Rightarrow 3y^2\frac{dy}{dx}=16 \\ \Rightarrow \frac{dy}{dx}=\frac{16}{3y^2} \\ \begin{array}{l}{m}_1{m}_2=-1\Rightarrow \left(\frac{-4x}{ky}\right)\left(\frac{16}{3y^2}\right)=-1\\ \Rightarrow 64x=3k{y}^3\\ \Rightarrow 64x=3k\left(16x\right)\Rightarrow k=\frac{4}{3}\end{array} \end{gathered}$$