Question

Through the vertex $O$ of a parabola $y^2=4x$, chords $OP$ and $OQ$ are drawn at right angles to one another. The locus of the middle point of $PQ$ is

• A. $y^2=2x+8$
• B. $y^2=x+8$
• C. $y^2=2x-8$
• D. None of these

Answer 1

Answer: (C)

$y^2=2x-8$

Given parabola is $y^2=4x$

Here $4a=4\Rightarrow a=1$ $...\left(1\right)$

Let $P\equiv ({t_1}^2,2t_1)$ and $Q\equiv ({t_2}^2,2t_2)$

Slope of $OP=\frac{2t_1}{{t_1}^2}=\frac{2}{t_1}$ and slope of $OQ=\frac{2}{t_2}$

Since $OP\perp OQ,$

$\therefore \frac{4}{t_1{t}_2}=-1\Rightarrow t_1{t}_2=-4$ $...\left(2\right)$

Let $R(\alpha ,\beta )$ be the middle point of $PQ$, then

$\alpha =\frac{{t_1}^2+{t_2}^2}{2}$ $...\left(3\right)$

and $\beta =t_1+t_2$ $...\left(4\right)$

From (4)

${\beta }^2={t_1}^2+{t_2}^2+2t_1{t}_2=2\alpha -8$

Hence locus of $R(\alpha ,\beta )$ is $y^2=2x-8$

Hence, option C.