Question

The locus of the mid point of the line segment joining the point $(4,3)$ and the points on the ellipse $x^2+2y^2=4$ is an ellipse with eccentricity :

• A. $\frac{\sqrt{3}}{2}$
• B. $\frac{1}{2\sqrt{2}}$
• C. $\frac{1}{\sqrt{2}}$
• D. $\frac{1}{2}$

Answer 1

Answer: (C)

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

Coordinate of $D$ is
$\left(\frac{2\cos \theta +4}{2},\frac{\sqrt{2}\sin \theta +3}{2}\right)\equiv (h,k)$
$\frac{2h-4}{2}=\cos \theta$ ...(i)
$\frac{2k-3}{\sqrt{2}}=\sin \theta$ ...(ii)
$(i{)}^2+(ii{)}^2$, then we get
${\left(\frac{2h-4}{2}\right)}^2+{\left(\frac{2k-3}{\sqrt{2}}\right)}^2=1$
$\Rightarrow \frac{(x-2{)}^2}{1}+\frac{{\left(y-\frac{3}{2}\right)}^2}{\left(\frac{1}{2}\right)}=1$
$\therefore$ Required eccentricity is
$e=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$