Let the line $L:\sqrt{2}x+y=\alpha$ pass through the point of intersection $P$ (in the first quadrant) of the circle $x^2+y^2=3$ and the parabola $x^2=2y$. Let $L$ touch two circles $C_1$ and $C_2$ of equal radius $2\sqrt{3}$. If the centres $Q_1$ and $Q_2$ of $C_1$ and $C_2$ lie on the $y$-axis, then the square of the area of the triangle $PQ_1Q_2$ is equal to
Step-by-Step Solution
Key Concept: Find $P$: from $x^2=2y$ and $x^2+y^2=3$: $y^2+2y-3=0\Rightarrow y=1$, $x=\sqrt{2}$. So $P=(\sqrt{2},1)$ and $\alpha=\sqrt{2}\cdot\sqrt{2}+1=3$. Centres $Q_1(0,\alpha_1)$, $Q_2(0,\alpha_2)$ on $y$-axis at distance $2\sqrt{3}$ from $L$: $|\alpha-3|/\sqrt{3}=2\sqrt{3}\Rightarrow\alpha=9$ or $-3$.
$P=(\sqrt{2},1)$, $Q_1=(0,9)$, $Q_2=(0,-3)$. Area of $\triangle PQ_1Q_2=\frac{1}{2}|Q_1Q_2|\cdot d(P,y\text{-axis})=\frac{1}{2}\cdot12\cdot\sqrt{2}=6\sqrt{2}$. $(\text{Area})^2=72$.
Correct Answer: 72