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

Question

Let the line $L:\sqrt2x+y=\alpha$ 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 the line $L$ touch two circles $C_1$ and $C_2$ of equal radius $2\sqrt3$. 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 $PQ_1Q_2$ is equal to \_\_\_\_\_\_\_\_.

Answer: 72

Accepted Answer

Substituting $x^2=2y$ into $x^2+y^2=3$: $y^2+2y-3=0\Rightarrow y=1$ (taking $y>0$), so $x=\sqrt2$: $P=(\sqrt2,1)$.

$L$ through $P$: $\sqrt2(\sqrt2)+1=\alpha\Rightarrow\alpha=3$: $L:\sqrt2x+y=3$.

Centres $(0,k)$ with $\dfrac{|k-3|}{\sqrt3}=2\sqrt3\Rightarrow|k-3|=6\Rightarrow k=9,-3$.

$$\text{Area}=\frac12(12)(\sqrt2)=6\sqrt2,\qquad (\text{Area})^2=72.$$

Solutions · 1