Mathematics - Ellipse Question with Solution | TestHub

Given, $\frac{x^2}{16}+\frac{y^2}{4}=1$ Here, $a=4,b=2$ Equation of normal is $$\begin{array}{rl}& 4x\sec \theta -2y\mathrm{cosec}\theta =12\\ & M\left(\frac{7\cos \theta }{2},\sin \theta \right)=(h,k)\text{ (say) }\\ & \therefore h=\frac{7\cos \theta }{2}\Rightarrow \cos \theta =\frac{2h}{7}\\ & \left[\because {\cos }^2\theta +{\sin }^2\theta =1\right]\\ & \end{array}$$ Hence, locus is $\frac{4x^2}{49}+y^2=1$
For given ellipse, $e^2=1-\frac{4}{16}=\frac{3}{4}$ $$\begin{array}{rc}\therefore e=\frac{\sqrt{3}}{2}& \\ \therefore x=\pm 4\times \frac{\sqrt{3}}{2}=\pm 2\sqrt{3}& \\ & [\because x=\pm ae]\dots \text{ (ii) }\end{array}$$ On solving Eqs. (i) and (ii), we get $$\begin{array}{rl}& \frac{4}{49}\times 12+y^2=1\Rightarrow y^2=1-\frac{48}{49}=\frac{1}{49}\\ & \Rightarrow y=\pm \frac{1}{7}\end{array}$$ $\therefore$ Required points $\left(\pm 2\sqrt{3},\pm \frac{1}{7}\right)$.