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Let $\mathrm{P Q}$ be a focal chord of the parabola $\mathrm{y^{2}=4 x}$ such that it subtends an angle of $\mathrm{\frac{\pi}{2}}$ at the point $(3,0)$ . Let the line segment PQ be also a focal chord of the ellipse $\mathrm{E}: \frac{x^{2}}{\mathrm{a}^{2}}+\frac{y^{2}}{\mathrm{~b}^{2}}=1, \mathrm{a}^{2}>\mathrm{b}^{2}.$ If e is the eccentricity of the ellipse $\mathrm{\mathrm{E}}$ , then the value of $\mathrm{\frac{1}{\mathrm{e}^{2}}}$ is equal to :
Option: 1
$1+\sqrt{2}$

Option: 2
$3+2 \sqrt{2}$

Option: 3
$1+2 \sqrt{3}$

Option: 4
$4+5 \sqrt{3}$

Answers (1)

best_answer

$\mathrm{Slope\: of \: P A=\frac{2 t_{1}}{t_{1}^{2}-3}}\\$

$\mathrm{Slope\: of\: Q A=\frac{2 t_{2}}{t_{2}^{2}-3}}$

Now these are perpendicular

$\mathrm{\Rightarrow \quad \frac{2 t_{1}}{t_{1}^{2}-3} \cdot \frac{2 t_{2}}{t_{2}^{2}-3}=-1} \\$

$\mathrm{\Rightarrow \quad 4 t_{1} t_{2}=-1\left(t_{1}^{2}-3\right)\left(t_{2}^{2}-3\right) }$

$\mathrm{Using \: t_{1} t_{2}=-1\, for\: focal\: chord}$

$\mathrm{\Rightarrow-4=-1\left(t_{1}^{2}-3\right)\left(\frac{1}{t_{1}^{2}}-3\right)} \\$

$\mathrm{\Rightarrow\left(t_{1}^{2}-3\right)\left(\frac{1}{t_{1}^{2}}-3\right)=4} \\$

$\mathrm{\Rightarrow 1-\frac{3}{t_{1}^{2}}-3 t_{1}^{2}+9=4}$

$\mathrm{\Rightarrow 3 t_{1}^{2}+\frac{3}{t_{1}^{2}}=6} \\$

$\mathrm{\Rightarrow 3\left(t_{1}^{2}+\frac{1}{t_{1}^{2}}-2\right)=0} \\$

$\mathrm{\Rightarrow\left(t_{1}-\frac{1}{t_{1}}\right)^{2}=0} \\$

$\mathrm{\Rightarrow t_{1}=\frac{1}{t_{1}}} \\$

$\mathrm{\Rightarrow t_{1}^{2}=1} \\$

$\mathrm{\Rightarrow t_{1}=1 \text { or }-1}$

$\mathrm{\therefore P(1,2), Q(1,-2)}$

$\mathrm{\therefore PQ}$ is also the focal chord and latus rectum of ellipse

$\mathrm{ P(1,2)}$ is end point ofn latus rectum

$\mathrm{\therefore a e=1\: and\: \frac{b^{2}}{a}=2}\\$

$\mathrm{a=\frac{1}{e}\: and\: b^{2}=2 a}$

$\mathrm{\text { Now } e^{2} =1-\frac{b^{2}}{a^{2}}} \\$

$\mathrm{e^{2} =1-\frac{2 a}{a^{2}} }$

$\mathrm{e^{2}=1-\frac{2}{a}} \\$

$\mathrm{e^{2}=1-2 e} \\$

$\mathrm{e^{2}+2 e-1=0} \\$

$\mathrm{e=\frac{-2+\sqrt{4+4}}{2}}$ ( negative rejected )

$\mathrm{=\frac{-2+2 \sqrt{2}}{2} }\\$

$\mathrm{=\sqrt{2}-1} \\$

$\mathrm{\therefore \frac{1}{e^{2}} =\frac{1}{(\sqrt{2}-1)^{2}}=\frac{1}{(3-2 \sqrt{2})} \cdot \frac{(3+2 \sqrt{2})}{(3+2 \sqrt{2})}} \\$

$\mathrm{=3+2 \sqrt{2}}$

Hence the correct answer is option 2.

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