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 $△OPQ$ is $3\sqrt{2}\text{sq. units}$, 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 $(1,\sqrt{2})$

Given parabola is $y^2=2x$
Let $P=(\frac{a^2}{2},a)$ and $Q=(\frac{b^2}{2},b)$


Slope of $OP,$
$m_1=\frac{2}{a}$
Slope of $OQ,$
$m_2=\frac{2}{b}$

Since $PQ$ is a diameter of the circle, so $\angle POQ={90}^{\circ }$
$\therefore m_1{m}_2=-1\Rightarrow ab=-4\cdots \left(1\right)$

Area of the triangle $△OPQ=3\sqrt{2}$
$\Rightarrow \frac{1}{2}\times OP\times OQ=3\sqrt{2}\Rightarrow \frac{1}{2}\times \sqrt{\frac{a^4}{4}+a^2}\times \sqrt{\frac{b^4}{4}+b^2}=3\sqrt{2}\Rightarrow \frac{1}{2}\times \frac{|ab|}{2\times 2}\times \sqrt{a^2+4}\times \sqrt{b^2+4}=3\sqrt{2}\Rightarrow (a^2+4)(b^2+4)=72\Rightarrow a^2+b^2=10\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right),$ we get
$a=\sqrt{2},2\sqrt{2}$
Thus, the coordinates of $P$ are $(1,\sqrt{2})\text{and}(4,2\sqrt{2})$