Question

It's JEE Advanced question

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Answer 1

Since P is an end point of the latus rectum of the parabola $y^2=4\lambda x,$

we have two positions for P, $(\lambda,\ 2\lambda)$ and $(\lambda,\ -2\lambda).$ Note that the focus is at a distance of $\lambda$ from the vertex.

Here I consider the point $P(\lambda,\ 2\lambda).$ Considering the other point also results the same answer.

Since the ellipse passes through this point, we have,

$\longrightarrow\dfrac{\lambda^2}{a^2}+\dfrac{4\lambda^2}{b^2}=1$

$\longrightarrow\dfrac{1}{a^2}+\dfrac{4}{b^2}=\dfrac{1}{\lambda^2}$

Now find the slope of tangent to the parabola at this point.

$\longrightarrow y^2=4\lambda x$

$\longrightarrow 2yy'=4\lambda$

$\longrightarrow y'=\dfrac{2\lambda}{y}$

At $P(\lambda,\ 2\lambda),$

$\longrightarrow y'=1$

So the slope of tangent to the ellipse should be -1 since the tangents are perpendicular to each other. [Product of slopes equal -1.]

Now find the slope of tangent to the ellipse at this point.

$\longrightarrow\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

$\longrightarrow\dfrac{2x}{a^2}+\dfrac{2yy'}{b^2}=0$

$\longrightarrow\dfrac{x}{a^2}+\dfrac{yy'}{b^2}=0$

At $P(\lambda,\ 2\lambda)$ where $y'=-1,$

$\longrightarrow\dfrac{\lambda}{a^2}-\dfrac{2\lambda}{b^2}=0$

$\longrightarrow\dfrac{a^2}{b^2}=\dfrac{1}{2}$

This means $a<b$ and so the major axis is along y axis. The ellipse is a vertical elipse.

Hence the eccentricity is given by,

$\longrightarrow e=\sqrt{1-\dfrac{a^2}{b^2}}$

$\longrightarrow e=\sqrt{1-\dfrac{1}{2}}$

$\longrightarrow\underline{\underline{e=\dfrac{1}{\sqrt2}}}$

Hence (A) is the answer.