The line x=8 is the directrix of the ellipse E:x^2a^2+y^2b^2=1 with corresponding focus (2…

Question

The line $x=8$ is the directrix of the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with corresponding focus $(2,0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through $(0,4\sqrt3)$ and intersects the $x$-axis at $Q$, then $(3\,PQ)^2$ is equal to \_\_\_\_\_\_.

Answer: 39

Accepted Answer

$\dfrac{a}e=8,\ ae=2\Rightarrow a^2=16\Rightarrow a=4,\ e=\dfrac12$; $b^2=a^2(1-e^2)=16\cdot\dfrac34=12$.

Tangent at $P(4\cos\theta,2\sqrt3\sin\theta)$: $\dfrac{x\cos\theta}4+\dfrac{y\sin\theta}{2\sqrt3}=1$. Through $(0,4\sqrt3)$: $\dfrac{4\sqrt3\sin\theta}{2\sqrt3}=1\Rightarrow\sin\theta=\dfrac12\Rightarrow\theta=30^\circ$.

$$P=(2\sqrt3,\sqrt3),\qquad Q=\left(\frac8{\sqrt3},0\right)\ (\text{setting }y=0).$$

$$PQ^2=\left(2\sqrt3-\frac8{\sqrt3}\right)^2+3=\frac43+3=\frac{13}3,\qquad (3PQ)^2=9\cdot\frac{13}3=39.$$

Solutions · 1