Let PQ be a focal chord of the parabola y^2=4x such that it subtends an angle of 2 at the …

Question

Let $PQ$ be a focal chord of the parabola $y^2=4x$ such that it subtends an angle of $\dfrac\pi2$ at the point $(3,0)$. Let the line segment $PQ$ also be a focal chord of the ellipse $E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, $a^2>b^2$. If $e$ is the eccentricity of $E$, then the value of $\dfrac1{e^2}$ is equal to:

Accepted Answer

As $$\angle PRQ = {\pi  \over 2}$$

$$\left( {{{{2 \over t}} \over {3 - {1 \over {{t^2}}}}}} \right)\,.\,\left( {{{ - 2t} \over {3 - {t^2}}}} \right) =  - 1$$

$$ \Rightarrow t =  \pm \,1$$

$$\therefore$$ $$P \equiv (1,2)$$ \& $Q(1, - 2)$

$$\therefore$$ for ellipse $${1 \over {{a^2}}} + {4 \over {{b^2}}} = 1$$ and $ae = 1$

$$ \Rightarrow {1 \over {{a^2}}} + {4 \over {{a^2}(1 - {e^2})}} = 1$$

$$ \Rightarrow 1 + {4 \over {(1 - {e^2})}} = {1 \over {{e^2}}}$$

$$ \Rightarrow (5 - {e^2}){e^2} = 1 - {e^2}$$

$$ \Rightarrow {e^4} - 6{e^2} + 1 = 0$$

$$ \Rightarrow {e^2} = {1 \over {3 - 2\sqrt 2 }} \Rightarrow {1 \over {{e^2}}} = 3 + 2\sqrt 2 $$

Solutions · 1