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  • Let a, b and be positive real numbers. Suppose P is an end point of the latus rectum of the parabola <img alt="

Let a, b and \lambda be positive real numbers. Suppose P is an end point of the latus rectum of the parabola \mathrm{y^2=4 \lambda x}, and suppose the ellipse \mathrm{\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

Option: 1

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


Option: 2

\frac{1}{{2}}


Option: 3

\frac{1}{{3}}


Option: 4

\frac{2}{{5}}


Answers (1)

best_answer

\mathrm{\text { We have, } y^2=4 \lambda x}

End point of latus rectum can be taken as \mathrm{P \equiv(\lambda, 2 \lambda)}
Now, \mathrm{y^2=4 \lambda x}

\mathrm{\Rightarrow 2 y \frac{d y}{d x}=4 \lambda \Rightarrow \frac{d y}{d x}=\frac{2 \lambda}{y}=\left[\frac{d y}{d x}\right]_{(\lambda, 2 \lambda)}=\frac{2 \lambda}{2 \lambda}=1}

Slope of tangent at P to the parabola = 1
 Slope of tangent at P to the ellipse = –1

\mathrm{\text { Also, we have, } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}

\mathrm{\begin{aligned} & \Rightarrow \frac{d y}{d x}=-\frac{2 x / a^2}{2 y / b^2}=-\frac{x b^2}{y a^2} \\ & \therefore \quad\left[\frac{d y}{d x}\right]_{(\lambda, 2 \lambda)}=-\frac{\lambda}{2 \lambda} \cdot \frac{b^2}{a^2}=-1 \Rightarrow \frac{a^2}{b^2}=\frac{1}{2} \\ & \therefore \quad e=\sqrt{1-\frac{a^2}{b^2}}=\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}} \end{aligned}}

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vishal kumar

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