Let the line L:√2x+y= pass through the point of the intersection P (in the first quadrant)…

Question

Let the line $L:\sqrt{2}x+y=\alpha $ pass through the point of the intersection $P$ (in the first quadrant) of the circle ${x}^{2}+{y}^{2}=3$ and the parabola ${x}^{2}=2y$. Let the line $L$ touch two circles ${C}_{1}$ and ${C}_{2}$ of equal radius $2\sqrt{3}$. If the centres ${Q}_{1}$ and ${Q}_{2}$ of the circles ${C}_{1}$ and ${C}_{2}$ lie on the $y$-axis, then the square of the area of the triangle ${PQ}_{1}{Q}_{2}$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 72

Accepted Answer

${x}^{2}+{y}^{2}=3$ and ${x}^{2}=2y$

${y}^{2}+2y-3=0$

$\Rightarrow (y+3)(y-1)=0$

$y=-3$ (Rejected) or $y=1$

For $y=1,x=\sqrt{2}\Rightarrow P(\sqrt{2},1)$

$p$ lies on the line

$\begin{gathered} \sqrt{2}x+y=\alpha \ \sqrt{2}(\sqrt{2})+1=\alpha \ \alpha =3 \end{gathered}$

For circle ${C}_{1}$

${Q}_{1}$ lies on $y$ axis

Let ${Q}_{1}(0,\alpha )$ coordinates

${R}_{1}=2\sqrt{3}$ (Given

Line $L$ act as tangent

Apply P $=r$ (condition of tangency)

$\begin{gathered} \Rightarrow |\frac{\alpha -3}{\sqrt{3}}|=2\sqrt{3} \ \Rightarrow |\alpha -3|=6 \end{gathered}$

$	herefore $ $\alpha -3=6$

$\Rightarrow \alpha =9$

$\begin{gathered} \alpha -3=-6 \ \Rightarrow \alpha =-3 \end{gathered}$

$\begin{gathered} 	riangleP{Q}_{1}{Q}_{2}=\frac{1}{2}|\begin{array}{ccc}\sqrt{2} \quad  1 \quad  1 \ 0 \quad  9 \quad  1 \ 0 \quad  -3 \quad  1\end{array}| \ =\frac{1}{2}(\sqrt{2}(12))=6\sqrt{2} \ {(	riangleP{Q}_{1}{Q}_{2})}^{2}=72 \end{gathered}$

Solutions · 1