Question

Let $P$ be the point $(1,0)$ and $Q$ a point of the locus $y^2=8x$. The locus of midpoint of $PQ$ is

• A. $x^2+4y+2=0$
• B. $x^2-4y+2=0$
• C. $y^2-4x+2=0$
• D. $y^2+4x+2=0$

Answer 1

The correct option is C $y^2-4x+2=0$

Let the point $Q$ be $(h,k)$

Then, $k^2=8h$

$\Rightarrow \frac{k^2}{8}=h$

Let the required midpoint be$(i,j)$

Then, $i=\frac{1+h}{2}$

$\Rightarrow i=\frac{1+\frac{k^2}{8}}{2}$

$\Rightarrow i=\frac{k^2+8}{16}$

$\Rightarrow 16i-8=k^2\to \left(1\right)$

and $j=\frac{k}{2}$

$\Rightarrow 2j=k\to \left(2\right)$

Using (1) and (2), we have

$16i-8=4j^2$

$\Rightarrow j^2-4i+2=0$

Replace $j$ by $y$ and $i$ by $x$.

Thus locus is: $y^2-4x+2=0$.