Let $\mathrm{a}, \mathrm{b}$ 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
Select the correct option:
A
$\frac{1}{\sqrt{2}}$
B
$\frac{1}{2}$
C
$\frac{1}{3}$
D
$\frac{2}{5}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$
\mathrm{y}^2=4 \lambda \mathrm{x}, \mathrm{P}(\lambda, 2 \lambda)
$
Slope of the tangent to the parabola at point P
$
\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{4 \lambda}{2 \mathrm{y}}=\frac{4 \lambda}{2 \mathrm{x} 2 \lambda}=1
$
Slope of the tangent to the ellipse at P
$
\frac{2 x}{a^2}+\frac{2 y y^{\prime}}{b^2}=0
$
As tangents are perpendicular $y^{\prime}=-1$
$
\begin{aligned}
& \Rightarrow \frac{2 \lambda}{\mathrm{a}^2}-\frac{4 \lambda}{\mathrm{~b}^2}=0 \Rightarrow \frac{\mathrm{a}^2}{\mathrm{~b}^2}=\frac{1}{2} \\
& \Rightarrow \mathrm{e}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}
\end{aligned}
$
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