The line x = 8 is the directrix of the ellips | vedclass

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(B) Given the directrix $x = \frac{a}{e} = 8$ and focus $(ae, 0) = (2, 0)$.From these,$ae = 2$ and $\frac{a}{e} = 8$.Multiplying gives $a^2 = 16$,so $

Vedclass · Accepted Answer

$39$

Vedclass · Answer

(B) Given the directrix $x = \frac{a}{e} = 8$ and focus $(ae, 0) = (2, 0)$.From these,$ae = 2$ and $\frac{a}{e} = 8$.Multiplying gives $a^2 = 16$,so $a = 4$. Then $e = \frac{2}{4} = \frac{1}{2}$.$b^2 = a^2(1 - e^2) = 16(1 - \frac{1}{4}) = 16(\frac{3}{4}) = 12$,so $b = 2\sqrt{3}$.The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{12} = 1$.The tangent at $P(4\cos	heta, 2\sqrt{3}\sin	heta)$ is $\frac{x\cos	heta}{4} + \frac{y\sin	heta}{2\sqrt{3}} = 1$.Since it passes through $(0, 4\sqrt{3})$,we have $\frac{0}{4} + \frac{4\sqrt{3}\sin	heta}{2\sqrt{3}} = 1$,which gives $2\sin	heta = 1$,so $\sin	heta = \frac{1}{2}$.Thus,$	heta = 30^\circ$. Point $P$ is $(4\cos 30^\circ, 2\sqrt{3}\sin 30^\circ) = (2\sqrt{3}, \sqrt{3})$.The tangent equation is $\frac{x\cos 30^\circ}{4} + \frac{y\sin 30^\circ}{2\sqrt{3}} = 1$ $\Rightarrow \frac{x\sqrt{3}}{8} + \frac{y}{4\sqrt{3}} = 1$.For $Q$,set $y = 0$: $\frac{x\sqrt{3}}{8} = 1 \Rightarrow x = \frac{8}{\sqrt{3}}$. So $Q = (\frac{8}{\sqrt{3}}, 0)$.$PQ^2 = (2\sqrt{3} - \frac{8}{\sqrt{3}})^2 + (\sqrt{3} - 0)^2 = (\frac{6-8}{\sqrt{3}})^2 + 3 = \frac{4}{3} + 3 = \frac{13}{3}$.$(3PQ)^2 = 9 	imes PQ^2 = 9 	imes \frac{13}{3} = 39$.

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