Question

Let $PQ$ be a chord of the parabola $y^2=4x$. A circle drawn with $PQ$ as a diameter passes through the vertex $V$ of the parabola. If area of $△PVQ=20$ sq. units, the the coordinates of $P$ are

• A. $(16,8)$
• B. $(16,-8)$
• C. $(-16,8)$
• D. $(-16,-8)$

Answer 1

Answer: (A), (B)

The correct options are
A $(16,8)$
B $(16,-8)$

$m$ of $PV=\frac{2t-0}{t^2-0}=\frac{2}{t}$
$\therefore$ the equation of $QV$ is $y=-\left(\frac{t}{2}\right)x$
Solving it with $y^2=4x,Q=(\frac{16}{t^2},-\frac{8}{t})$
Now, area of $△PVQ=\frac{1}{2}PV.VQ=20$ (given)
$\therefore {PV}^2.{VQ}^2={40}^2\Rightarrow ({\left(t^2\right)}^2+{\left(2t\right)}^2)({\left(\frac{16}{t^2}\right)}^2+(-\frac{8}{t}))={40}^2$
$\Rightarrow t^2(4+t^2)\left(\frac{1}{t^2}\right)(\frac{256}{t^2}+64)={40}^2$
$\Rightarrow \frac{256\times 4}{t^2}+256+256+64t^2={40}^2$
$\Rightarrow (t^2-16)(t^2-1)=0\Rightarrow t=\pm 4,\pm 1$