Let PQ be a focal chord of y2=4x such that it subtends an angle of π2 at the point (3,0). Let the line segment PQ is also focal chord of ellipse x2a2+y2b2=1,a>b If e is eccentricity of ellipse, then 1e2 is ( where . represents Greatest integer function)

(Slope of PA )(slope of QA)=−1⇒t=1Then P and Q are ends of LLRAnd P(1,2) Q(1,-2)LLR of ellipse is 2b2a=4 and ae=1a21−e2=2a⇒a2−2a−1=0a=1+2Then 1e2=3+22

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Let PQ be a focal chord of $y^{2} = 4 x$ such that it subtends an angle of $\frac{π}{2}$ at the point (3,0). Let the line segment PQ is also focal chord of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b$ If e is eccentricity of ellipse, then $[\frac{1}{e^{2}}] i s$ ( where $[.]$ represents Greatest integer function)

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Detailed Solution

(Slope of PA )(slope of $QA) = - 1 \Rightarrow t = 1$
Then P and Q are ends of LLR
And P(1,2) Q(1,-2)
LLR of ellipse is $\frac{2 b^{2}}{a} = 4 and ae = 1$
$\begin{array}{l} a^{2} (1 - e^{2}) = 2 a \\ \Rightarrow a^{2} - 2 a - 1 = 0 \\ a = 1 + \sqrt{2} \end{array}$
Then $\frac{1}{e^{2}} = 3 + 2 \sqrt{2}$
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