Question

Let $P$ be the point $(1,0)$ and $Q$ a point on the curve $y^2=8x$. The locus of mid-point 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

Answer: (D)

$y^2-4x+2=0$

The coordinates of $P$ are $(1,0)$.

A general point $Q$ on $y^2=8x$ is $(h,k)$.

such that $k^2=8h$

Mid-point of $PQ$ is $(\alpha ,\beta )$.

Therefore, $\alpha =\frac{h+1}{2}\Rightarrow 2\alpha =h+1$ ....(i)
and $\beta =\frac{k+0}{2}\Rightarrow 2\beta =k$
Then, from equation (i), we get
$(2\beta {)}^2=8(2\alpha -1)$
$\Rightarrow {\beta }^2=4\alpha -2$
So, the locus of $(\alpha ,\beta )$ is $y^2-4x+2=0$.