The normal at a point P on the ellipse x^2+4 y^2=16 meets the x...

The normal at a point

P

on the ellipse

x^2+4 y^2=16

meets the

x

-axis at

Q

. If

M

is the mid-point of the line segment

P Q

, then the locus of

M

intersects the latusrectum of the given ellipse at the points The normal at a point P on the ellipse x^2+4 y^2=16 meets the x -axis at Q . If M is the mid-point of the line segment

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

Options

A $\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{2}{7}\right)$
B $\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4}\right)$
C $\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$
D $\left(\pm 2 \sqrt{3}, \pm \frac{4 \sqrt{3}}{7}\right)$

Correct answer

C. $\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$

Step-by-step solution

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<img src="/api/proxy-image?url=https%3A%2F%2Ffirebasestorage.googleapis.com%2Fv0%2Fb%2Fquantrexacademy-app.firebasestorage.app%2Fo%2Fquestions%252Ffigs%252Fpyq%252Fjee_advanced%252Fh5Ha-7vIkDAyJhn9AIK2wVfBtoXgLlrlSF9VVmEZDp0.original.fullsize.png%3Falt%3Dmedia&clean=1&v=wm2" data-qx-orig-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2Fh5Ha-7vIkDAyJhn9AIK2wVfBtoXgLlrlSF9VVmEZDp0.original.fullsize.png?alt=media" data-qx-storage-src="https://firebasestorage.googleapis.com/v0/b/quantrexacademy-app.firebasestorage.app/o/questions%2Ffigs%2Fpyq%2Fjee_advanced%2Fh5Ha-7vIkDAyJhn9AIK2wVfBtoXgLlrlSF9VVmEZDp0.original.fullsize.png?alt=media">

\\ For given ellipse,

e^2=1-\frac{4}{16}=\frac{3}{4}

\begin{array}{rc} \therefore \quade=\frac{\sqrt{3}}{2}\\ \therefore \quad x=\pm4 \times \frac{\sqrt{3}}{2}=\pm2 \sqrt{3}\\ & {[\because x=\pma e] \ldots \text{ (ii) }} \end{array}

On solving Eqs. (i) and (ii), we get

\begin{aligned}& \frac{4}{49} \times12+y²=1 \Rightarrow y²=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)

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