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 DOPQ is 3 \textrm{ } \sqrt{2} then which of the following is(are) the coordinates of P?

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Question

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 DOPQ is $3 \sqrt{2}$ then which of the following is(are) the coordinates of P?

$(1, \sqrt{2})$

$(4, 2 \sqrt{2})$

$(\frac{1}{4}, \frac{1}{\sqrt{2}})$

$(9, 3 \sqrt{2})$

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$m_{OP} = \frac{2}{t_{1}}; m_{OQ} = \frac{2}{t_{2}}$

m_OP · m_OQ = – 1 $\Rightarrow \frac{2}{t_{1}} \cdot \frac{2}{t_{2}}$ = – 1 ⇒ t₁t₂ = – 4

Area ( $∆$ OPQ) = 3

$\frac{1}{2} | \begin{matrix} {at}_{1}^{2} & 2 {at}_{1} & 1 \\ {at}_{2}^{2} & 2 {at}_{2} & 1 \\ 0 & 0 & 1 \end{matrix} | = \pm 3 \sqrt{3}$

${2a}^{2} (t_{1}^{2} t_{2} - t_{1} t_{2}^{2}) = \pm 6 \sqrt{2}$

$t_{1} t_{2} (t_{1} - t_{2}) = \pm 12 \sqrt{2}$

$\Rightarrow t_{1} - t_{2} = \pm 3 \sqrt{2}$

⇒ (t₁ + t₂)² = (t₁ – t₂)² + 4_t1t₂ = 18 – 16 = 2

⇒ t₁ + t₂ = ± $\sqrt{2}$ …….(2)

From equation (1) & (2)

$\Rightarrow t_{1} = 2 \sqrt{2} or \sqrt{2}$

$∴$ Coordinates of P are $(4, 2 \sqrt{2}) or (1, \sqrt{2}) .$