Let PQ be a focal chord of the parabola y^{2} | vedclass

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(B) Let the coordinates of $P$ be $(t^{2}, 2t)$ and $Q$ be $(\frac{1}{t^{2}}, -\frac{2}{t})$.Since $PQ$ subtends an angle of $\frac{\pi}{2}$ at $R(3,

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$3+2\sqrt{2}$

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(B) Let the coordinates of $P$ be $(t^{2}, 2t)$ and $Q$ be $(\frac{1}{t^{2}}, -\frac{2}{t})$.Since $PQ$ subtends an angle of $\frac{\pi}{2}$ at $R(3, 0)$,the product of the slopes of $PR$ and $QR$ is $-1$.Slope of $PR = \frac{2t-0}{t^{2}-3} = \frac{2t}{t^{2}-3}$.Slope of $QR = \frac{-2/t-0}{1/t^{2}-3} = \frac{-2/t}{(1-3t^{2})/t^{2}} = \frac{-2t}{1-3t^{2}}$.Since the product is $-1$,we have $\frac{2t}{t^{2}-3} 	imes \frac{-2t}{1-3t^{2}} = -1$.$\frac{-4t^{2}}{(t^{2}-3)(1-3t^{2})} = -1 \Rightarrow 4t^{2} = (t^{2}-3)(1-3t^{2}) = t^{2} - 3t^{4} - 3 + 9t^{2} = -3t^{4} + 10t^{2} - 3$.$3t^{4} - 6t^{2} + 3 = 0 \Rightarrow 3(t^{2}-1)^{2} = 0 \Rightarrow t^{2} = 1$.Thus,$P$ is $(1, 2)$ and $Q$ is $(1, -2)$.The length of the chord $PQ$ is $4$,which is the length of the latus rectum of the ellipse $E$. Thus,$\frac{2b^{2}}{a} = 4 \Rightarrow b^{2} = 2a$.The focus of the ellipse is $(ae, 0)$. Since $PQ$ is a focal chord,the line $x=1$ must pass through the focus $(ae, 0)$,so $ae = 1$.Using $b^{2} = a^{2}(1-e^{2})$,we substitute $b^{2} = 2a$ and $e^{2} = \frac{1}{a^{2}}$:$2a = a^{2}(1 - \frac{1}{a^{2}}) = a^{2} - 1$.$a^{2} - 2a - 1 = 0$. Solving for $a$,$a = \frac{2 \pm \sqrt{4+4}}{2} = 1 \pm \sqrt{2}$. Since $a>0$,$a = 1+\sqrt{2}$.Then $e^{2} = \frac{1}{a^{2}} = \frac{1}{(1+\sqrt{2})^{2}} = \frac{1}{1+2+2\sqrt{2}} = \frac{1}{3+2\sqrt{2}} = 3-2\sqrt{2}$.Therefore,$\frac{1}{e^{2}} = \frac{1}{3-2\sqrt{2}} = 3+2\sqrt{2}$.

List-$I$	List-$II$
$(A)$ $\left[\frac{p}{2}\left(t+\frac{1}{t}\right), \frac{q}{2}\left(t-\frac{1}{t}\right)\right]$	$(I)$ parabola
$(B)$ $(p+q \cos \theta, r+q \sin \theta)$	$(II)$ circle
$(C)$ $(p+\lambda^2, q-\lambda)$	$(III)$ ellipse
	$(IV)$ hyperbola

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