Given, $\quad \frac{x^2}{16}+\frac{y^2}{4}=1$
Here, $\quad a=4, b=2$
Equation of normal is
$$
\begin{aligned}
& 4 x \sec \theta-2 y \operatorname{cosec} \theta=12 \\
& M\left(\frac{7 \cos \theta}{2}, \sin \theta\right)=(h, k) \text { (say) } \\
& \therefore \quad h=\frac{7 \cos \theta}{2} \Rightarrow \cos \theta=\frac{2 h}{7} \\
& {\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]} \\
&
\end{aligned}
$$
Hence, locus is $\frac{4 x^2}{49}+y^2=1$


For given ellipse, $e^2=1-\frac{4}{16}=\frac{3}{4}$
$$
\begin{array}{rc}
\therefore \quad e=\frac{\sqrt{3}}{2} \\
\therefore \quad x=\pm 4 \times \frac{\sqrt{3}}{2}=\pm 2 \sqrt{3} \\
& {[\because x=\pm a e] \ldots \text { (ii) }}
\end{array}
$$
On solving Eqs. (i) and (ii), we get
$$
\begin{aligned}
& \frac{4}{49} \times 12+y^2=1 \Rightarrow y^2=1-\frac{48}{49}=\frac{1}{49} \\
& \Rightarrow \quad y=\pm \frac{1}{7}
\end{aligned}
$$
$\therefore$ Required points $\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$.