Let S be the focus of the parabola y2 = 8x and let PQ be the common chord of the circle x2 + y2 – 2x – 4y = 0 and the given parabola. The area of the triangle PQS, is

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Question

Let S be the focus of the parabola y² = 8x and let PQ be the common chord of the circle x² + y² – 2x – 4y = 0 and the given parabola. The area of the triangle PQS, is

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Parabola $\Rightarrow$ y² = 8x focus = (2, 0)

Circle $\Rightarrow$ x² + y² – 2x – 4y = x(x – 2) + y(y – 4) = 0

$∴$ circle with diametric ends (0, 0) and (2, 4)

which are vertex and one end of Latus rectum.

$∴$ Area of DPQS

$= \frac{1}{2} \times 2 \times 4 = 4$ Ans.

Aliter: Since parabola and circle both passes through origin.

$∴$ one common point P (0, 0).

Solving above two equations simultaneously, x² + 6x – 8 $\sqrt{2 x}$ = 0

By hit and trail, x = 2 satisfy above equation.

So another common point Q(2, 4)

$∴$ Area of $∆$ PQS = $\frac{1}{2} \times 2 \times 4$ = 4 Ans.