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 $P$? (1) $(4, 2\sqrt{2})$ (2) $(9, 3\sqrt{2})$ (3) $\left(\frac{1}{4}, \frac{1}{\sqrt{2}}\right)$ (4) $(1, \sqrt{2})$

Let P and Q be distinct points on the parabola y² = 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 Δ OPQ is 3√2, then which of the following is (are) the coordinates of P?

$y^{2}=2x,\quad P,Q \in \text{parabola}, P \neq Q$

$\text{circle with diameter } PQ \text{ passes through } O$

P ∈ Q1, [triangle OPQ] = 3√2

$\text{To find: coordinates of } P$

$y^{2}=2x=4ax \Rightarrow a=\dfrac{1}{2}$

$P=\left(\dfrac{t_1^{2}}{2},t_1\right),\quad Q=\left(\dfrac{t_2^{2}}{2},t_2\right)$

$\text{Circle with diameter } PQ \text{ through } O$

∠ POQ = 90^∘ ⇒ OP⃗·OQ⃗=0

$\left(\dfrac{t_1^{2}}{2}\right)\left(\dfrac{t_2^{2}}{2}\right)+t_1 t_2=0$

$t_1 t_2\left(\dfrac{t_1 t_2}{4}+1\right)=0$

t₁ t₂=0 ⇒ degenerate triangle OPQ , rejected

$\therefore\ t_1 t_2=-4$

[triangle OPQ]=1/2|x₁ y₂-x₂ y₁|

$=\dfrac{1}{2}\left|\dfrac{t_1 t_2}{2}(t_1-t_2)\right|=|t_1-t_2|$

$\Rightarrow\ |t_1-t_2|=3\sqrt{2}$

$t_1 t_2=-4 \Rightarrow t_2=-\dfrac{4}{t_1}$

$t_1-t_2=t_1+\dfrac{4}{t_1}=3\sqrt{2}\quad (t_1>0)$

$t_1^{2}-3\sqrt{2}\,t_1+4=0$

$D=18-16=2,\quad t_1=\dfrac{3\sqrt{2}\pm\sqrt{2}}{2}$

$t_1=2\sqrt{2}\ \text{or}\ t_1=\sqrt{2}$

$P=(4,2\sqrt{2})\ \text{or}\ P=(1,\sqrt{2})$

$\text{Matches options } (1) \text{ and } (4)$