Let the line L: sqrt{2}x + y = alpha pass thr | vedclass

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(B) Given equations: $x^2 + y^2 = 3$ and $x^2 = 2y$.Substituting $x^2 = 2y$ into the circle equation: $2y + y^2 = 3 \Rightarrow y^2 + 2y - 3 = 0$.$(y

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(B) Given equations: $x^2 + y^2 = 3$ and $x^2 = 2y$.Substituting $x^2 = 2y$ into the circle equation: $2y + y^2 = 3 \Rightarrow y^2 + 2y - 3 = 0$.$(y + 3)(y - 1) = 0 \Rightarrow y = 1$ (since $P$ is in the first quadrant,$y > 0$).For $y = 1$,$x^2 = 2(1) = 2 \Rightarrow x = \sqrt{2}$. Thus,$P = (\sqrt{2}, 1)$.Since $P$ lies on $L: \sqrt{2}x + y = \alpha$,we have $\sqrt{2}(\sqrt{2}) + 1 = \alpha \Rightarrow \alpha = 3$.Line $L$ is $\sqrt{2}x + y - 3 = 0$. The centres $Q_1, Q_2$ lie on the $y$-axis,so let $Q = (0, k)$.The radius of the circles is $r = 2\sqrt{3}$. The distance from $(0, k)$ to $\sqrt{2}x + y - 3 = 0$ is $r$:$\frac{|\sqrt{2}(0) + k - 3|}{\sqrt{(\sqrt{2})^2 + 1^2}} = 2\sqrt{3}$ $\Rightarrow \frac{|k - 3|}{\sqrt{3}} = 2\sqrt{3}$ $\Rightarrow |k - 3| = 6$.$k - 3 = 6 \Rightarrow k = 9$ or $k - 3 = -6 \Rightarrow k = -3$.So,$Q_1 = (0, 9)$ and $Q_2 = (0, -3)$.The area of $	riangle PQ_1Q_2$ with vertices $(\sqrt{2}, 1), (0, 9), (0, -3)$ is:$	ext{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = \frac{1}{2} |\sqrt{2}(9 - (-3)) + 0 + 0| = \frac{1}{2} |\sqrt{2}(12)| = 6\sqrt{2}$.The square of the area is $(6\sqrt{2})^2 = 36 	imes 2 = 72$.

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