The line x = 8 is the directrix of the ellipse E : x 2 a 2 + 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 ( 0 , 4 3 ) and intersects the x -axis at Q ,

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Solution

Q.
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 $(0, 4 \sqrt{3})$ and intersects the $x$ -axis at $Q$ , then ${(3 P Q)}^{2}$ is equal to __________ . [2023]

Ans.
(39)

Given, the line $x = 8$ is the directrix of the ellipse

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, \frac{a}{e} = 8$ ...(i) and focus, $a e = 2$ ...(ii)

From (i) and (ii)

$8 e = \frac{2}{e} \Rightarrow e^{2} = \frac{1}{4}$ $∴$ $e = \frac{1}{2}$

and $a = 4, b^{2} = a^{2} (1 - e^{2}) \Rightarrow b^{2} = 16 (\frac{3}{4}) = 12$

Now, equation of tangent at $P (4 \cos θ, 2 \sqrt{3} \sin θ)$ is

$\frac{x \cos θ}{4} + \frac{y \sin θ}{2 \sqrt{3}} = 1$ and it passes through $(0, 4 \sqrt{3})$ , so it satisfies the tangent equation.

$0 + \frac{4 \sqrt{3} \sin θ}{2 \sqrt{3}} = 1 \Rightarrow \sin θ = \frac{1}{2}$ $∴$ $θ = 30 °$

$∴$ $Point P = (2 \sqrt{3}, \sqrt{3}) and Q (\frac{8}{\sqrt{3}}, 0)$

$P Q^{2} = {(2 \sqrt{3} - \frac{8}{\sqrt{3}})}^{2} + {(\sqrt{3} - 0)}^{2} = \frac{13}{3}$

$∴$ ${(3 P Q)}^{2} = 9 \times P Q^{2} = 9 \times \frac{13}{3} = 39$

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