Question

Let $P$ and $Q$ be distinct points on the parabola $y^2=2x$ such that a circle with $PQ$ as diameter passes through the vertex $O$ of the parabola. If $P$ lies in the first quadrant and the area of the triangle $\Delta OPQ$ is $3\sqrt{2}$, then which of the following is (are) the coordinates of $P$?

• A. $(4,2\sqrt{2})$
• B. $(9,3\sqrt{2})$
• C. $(\frac{1}{4},\frac{1}{\sqrt{2}})$
• D. $(1,\sqrt{2})$

Answer 1

Answer: (A), (D)

The correct options are
A $(4,2\sqrt{2})$
D $(1,\sqrt{2})$

Parabola $:y^2=2x$
A point at parabola be $P(\frac{t_1^2}{2},t_1),Q(\frac{t_2^2}{2},t_2)$
A circle is drawn with $PQ$ as diameter and passes through $O(0,0)$.
$\angle POQ={90}^o$ in $△OPQ$
$\because PO$ and $OQ$ are perpendicular.
$\therefore ($Slope of $OP)\times ($Slope of $OQ)=-1$
$\Rightarrow \frac{2}{t_1}\times \frac{-2}{t_2}=-1\Rightarrow t_1{t}_2=4$..............{1}
Area of $△OPQ=\frac{1}{2}\left|\begin{array}{ccc}\frac{t_1^2}{2}& t_1& 1\\ \frac{t_2^2}{2}& -t_2& 1\\ 0& 0& 1\end{array}\right|=3\sqrt{2}$
$\Rightarrow t_1+t_2=3\sqrt{2}$.......... {2}
From {1} and {2}, we get
$(t_1,t_2)$ either $(2\sqrt{2},\sqrt{2})$ or $(\sqrt{2},2\sqrt{2})$
$\therefore$ Coordinates of $P$ are $(4,2\sqrt{2})$ or $(1,\sqrt{2})$
Hence, option A,D.