Question

Let $PQ$ be a chord of the parabola $y^2=8x$. A circle drawn with $PQ$ as diameter passes through the vertex $V$ of the parabola. If area of $\Delta PVQ=80$ square unit, then the coordinates of $P$ are

• A. $(32,-16)$
• B. $(-32,16)$
• C. $(-32,-16)$
• D. $(32,16)$

Answer 1

Answer: (A), (D)

The correct options are
A $(32,-16)$
D $(32,16)$

Let $P(2t_1^2,4t_1)$ and $Q(2t_2^2,4t_2)$
Slope of $VP=\frac{4t_1-0}{2t_1^2-0}={\frac{2}{t}}_1$
Slope of $VQ=\frac{4t_2-0}{2t_2^2-0}={\frac{2}{t}}_2$

Diameter subtends ${90}^{\circ }$ at the circumference
Slope of $VP\times$Slope of $VQ=-1$
$\Rightarrow t_1{t}_2=-\mathrm{4....}\left(1\right)$

Area of $\Delta PQV=80$
$\frac{1}{2}\left|\begin{array}{ccc}2t_1^2& 4t_1& 1\\ 2t_2& 4t_2^2& 1\\ 0& 0& 1\end{array}\right|=80\Rightarrow |8t_1^2{t}_2-8t_2^2{t}_1|=160\Rightarrow |t_1-t_2|=\mathrm{5...}\left(2\right)$

From the equation $\left(1\right)$ and $\left(2\right)$
$t_1=\pm 1\text{or}\pm 4$
The points $P$ can take $(2,\pm 4)$ and $(32,\pm 16)$