Let a,b and λ be positive real numbers. Suppose P is an end point of the latus rectum of the parabola y2=4λx, and suppose the ellipse x2a2+y2b2=1,(b>a) 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

y2=4λx,P(λ,2λ)Slope of the tangent to the parabola at point Pdydx=4λ2y=4λ2×2λ=1As tangents are perpendicular y′=−1Slope of the tangent to the ellipse at P ⇒2λa2−4λb2=0⇒a2b2=122xa2+2yy′b2=0 ⇒e=1−12=12

Let a,b and λ be positive real numbers. Suppose P is an end point of the latus rectum of the parabola y2=4λx, and suppose the ellipse x2a2+y2b2=1,(b>a) 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

y2=4λx,P(λ,2λ)Slope of the tangent to the parabola at point Pdydx=4λ2y=4λ2×2λ=1As tangents are perpendicular y′=−1Slope of the tangent to the ellipse at P ⇒2λa2−4λb2=0⇒a2b2=122xa2+2yy′b2=0 ⇒e=1−12=12

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Let a,b and $λ$ be positive real numbers. Suppose P is an end point of the latus rectum of the parabola $y^{2} = 4 λx$ , and suppose the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (b > a)$ 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

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

$\frac{2}{5}$

$\frac{1}{2}$

$\frac{1}{3}$

answer is A.

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Detailed Solution

$y^{2} = 4 λx, P (λ, 2 λ)$
Slope of the tangent to the parabola at point P
$\frac{dy}{dx} = \frac{4 λ}{2 y} = \frac{4 λ}{2 \times 2 λ} = 1$
As tangents are perpendicular $y^{'} = - 1$
Slope of the tangent to the ellipse at P
$\Rightarrow \frac{2 λ}{a^{2}} - \frac{4 λ}{b^{2}} = 0 \Rightarrow \frac{a^{2}}{b^{2}} = \frac{1}{2}$
$\frac{2 x}{a^{2}} + \frac{2 {yy}^{'}}{b^{2}} = 0 \Rightarrow e = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt{2}}$