The line $x = 8$ is the directrix of the ellipse $E: \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with the corresponding focus $(2, 0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $\left(0, 4\sqrt{3}\right)$ and intersects the x-axis at Q, then $(3PQ)^{2}$ is equal to ____.
Step-by-Step Solution
Key Concept: From directrix $x = a/e = 8$ and focus $ae = 2$: multiply to get $a^2 = 16$, $a = 4$, $e = 1/2$, $b^2 = 12$. Write tangent at $P(4\cos\theta, 2\sqrt{3}\sin\theta)$ and use the condition it passes through $(0, 4\sqrt{3})$.
$a/e = 8$, $ae = 2 \Rightarrow a^2 = 16$, $e = 1/2$, $b^2 = 12$. Tangent at $P(4\cos\theta, 2\sqrt{3}\sin\theta)$: $\frac{x\cos\theta}{4} + \frac{y\sin\theta}{2\sqrt{3}} = 1$. Passes through $(0,4\sqrt{3})$: $\frac{4\sqrt{3}\sin\theta}{2\sqrt{3}} = 1 \Rightarrow 2\sin\theta = 1 \Rightarrow \theta = 30°$. $P = (2\sqrt{3},\sqrt{3})$, $Q = (8/\sqrt{3}, 0)$. $PQ^2 = (2\sqrt{3}-8/\sqrt{3})^2 + 3 = (2/\sqrt{3})^2 + 3 = 4/3 + 3 = 13/3$. $(3PQ)^2 = 9 \cdot 13/3 = 39$.
Correct Answer: 39