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Let $a , b$ and $\lambda$ be positive real numbers. Suppose $P$ is an end point of the latus rectum of the parabola $y^2=4 \lambda x$, and suppose the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ passes through the point $P$. If the tangents to the parabola and the ellipse at the point $P$ are perpendicular to each other, then the eccentricity of the ellipse is
$\frac{1}{\sqrt{2}}$
$\frac{1}{2}$
$\frac{1}{3}$
$\frac{2}{5}$
Solution
$y ^2=4 \lambda x , P (\lambda, 2 \lambda)$
Slope of the tangent to the parabola at point $P$
$\frac{ dy }{ dx }=\frac{4 \lambda}{2 y }=\frac{4 \lambda}{2 x 2 \lambda}=1$
Slope of the tangent to the ellipse at $P$
$\frac{2 x }{ a ^2}+\frac{2 yy ^{\prime}}{ b ^2}=0$
As tangents are perpendicular $y^{\prime}=-1$
$\Rightarrow \frac{2 \lambda}{a^2}-\frac{4 \lambda}{b^2}=0 \Rightarrow \frac{a^2}{b^2}=\frac{1}{2}$
$\Rightarrow e=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$
