Question

Let P be the point (2,4) and Q is a point on the locus $x^2+2y^2=4$, then locus of mid point of PQ is

• A. circle
• B. parabola
• C. ellipse
• D. hyperbola

Answer 1

The correct option is C ellipse

Given curve is $x^2+2y^2=4\Rightarrow \frac{x^2}{2^2}+\frac{y^2}{(\sqrt{2}{)}^2}=1$
Thus any point on this curve is taken as, $Q\equiv (2\cos \theta ,\sqrt{2}\sin \theta )$
Let midpoint of $PQ$ be $R(h,k)$
$\Rightarrow h=\frac{2\cos \theta +2}{2}=1+\cos \theta \Rightarrow \cos \theta =h-1..\left(1\right)$
and $k=\frac{4+\sqrt{2}\sin \theta }{2}\Rightarrow \sin \theta =\sqrt{2}(k-2)...\left(2\right)$
Now Squaring and adding (1) and (2) we get,
$(h-1{)}^2+2(k-2{)}^2={\cos }^2\theta +{\sin }^2\theta =1$, using trigonometric identity
Thus locus of $R(h,k)$ is, $(x-1{)}^2+2(y-2{)}^2=1$, which is clearly an ellipse.