$\text{ The normal at a point }P\text{ on the ellipse }{x}^2+4y^2=16\text{ meets the }x\text{ -axis at }Q\text{ . If }M\text{ is the midpoint of the line segment }$

$PQ\text{ , then the locus of }M\text{ intersects the latus rectums of the given ellipse at points }$

$\text{ Normal at }P\text{ is given by }4x\sec \varphi -2y\mathrm{cosec}\varphi =12$

$\begin{array}{l}\therefore Q\equiv (3\cos \varphi ,0)\\ \text{ Let mid-point of }PQ\text{ be }M(\alpha ,\beta )\text{ . }\\ \therefore \alpha =\frac{3\cos \varphi +4\cos \varphi }{2}=\frac{7}{2}\cos \varphi \\ \text{ or }\cos \varphi =\frac{2}{7}\alpha \\ \text{ and }\beta =\sin \varphi \end{array}$

$\text{ Using }{\cos }^2\varphi +{\sin }^2\varphi =1\text{ , we have }$

$\begin{array}{l}\frac{4}{49}{\alpha }^2+{\beta }^2=1\\ \text{ or }\frac{4}{49}{x}^2+y^2=1----\left(1\right)\end{array}$

$\begin{array}{l}\text{ Now, the latus rectum to above ellipse is }\\ x=\pm 2\sqrt{3}-----\left(2\right)\end{array}$

$\text{ Solving (1) and (2), we have }$

$\begin{array}{r}\frac{48}{49}+y^2=1\\ \text{ or }y=\pm \frac{1}{7}\end{array}$

$\text{ The points of intersection are }(\pm 2\sqrt{3},\pm 1/7)\text{ . }$