First slide

Ellipse

Question

$Let a, b and λ be positive real numbers. Suppose P is an end point of the latus rectum of the$
$parabola y^{2} = 4 λ x, and suppose the ellipse \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 passes through the point P . If the tangents to$
$the parabola and the ellipse at the point P are perpendicular to each other, then the eccentricity of the ellipse is$

Moderate

A

$\frac{1}{\sqrt{2}}$
B

$\frac{1}{2}$
C

$\frac{1}{3}$
D

$\frac{2}{5}$

Solution

$\begin{array}{l} y^{2} = 4 λ x, P (λ, 2 λ) \\ Slope of the tangent to the parabola at point P \end{array}$
$\begin{array}{l} \frac{d y}{d x} = \frac{4 λ}{2 y} = \frac{4 λ}{2 (2 λ)} = 1 \\ Slope of the tangent to the ellipse at P \\ \frac{2 x}{a^{2}} + \frac{2 y y^{'}}{b^{2}} = 0 \end{array}$
$\begin{array}{l} As tangents are perpendicular, [y^{'}] (1) = - 1 \Rightarrow y' = - 1 \\ \Rightarrow \frac{2 λ}{a^{2}} - \frac{4 λ}{b^{2}} = 0 \Rightarrow \frac{a^{2}}{b^{2}} = \frac{1}{2} (a t t h e p o i n t (λ, 2 λ)) \\ \Rightarrow e = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}} \end{array},$

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