Question

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

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

Answer 1

The correct option is C $(\pm 2\sqrt{3},\pm \frac{1}{7})$

Given Ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$

$e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2}$

$\because P$ is a point on the ellipse

So, $P=(4\cos \theta ,2\sin \theta )$

Equation of normal to the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ at point $(x_1,y_1)=(4\cos \theta ,2\sin \theta )$ is given by

$a^2{y}_1(x-x_1)=b^2{x}_1(y-y_1)$

$⟹16\times 2\sin \theta (x-4\cos \theta )=4\times 4\cos \theta (y-2\sin \theta )$

$⟹2x\sin \theta -8\sin \theta \cos \theta =y\cos \theta -2\sin \theta \cos \theta$

$⟹2x\sin \theta =y\cos \theta +6\sin \theta \cos \theta$

$⟹\frac{2x}{\cos \theta }=\frac{y}{\sin \theta }+6$

$⟹2x\sec \theta -y\text{cosec}\theta =6$

It meet the x-axis at $Q(3\cos \theta ,0)$

$\therefore M=(\frac{7}{2}\cos \theta ,\sin \theta )=(x,y)$

Locus of $M$ is

$\frac{x^2}{{\left(\frac{7}{2}\right)}^2}+\frac{y^2}{1}=1$

Latus rectum of the given ellipse is
$x=\pm ae=\pm \sqrt{16-4}=\pm 2\sqrt{3}$
So locus of $M$ meets the latus rectum at points for which
$y^2=1-\frac{12\times 4}{49}=\frac{1}{49}$ $\Rightarrow$ $y=\pm \frac{1}{7}$
Hence, the required point is $(\pm 2\sqrt{3},\pm \frac{1}{7})$.