The ellipse frac{x^2}{a^2}+frac{y^2}{b^2}=1 ( | vedclass

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(A) Let the point of intersection be $(x_0, y_0)$. For the parabola $y^2=4ax$,the slope of the tangent $m_1$ at $(x_0, y_0)$ is given by $2y y' = 4a \

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(A) Let the point of intersection be $(x_0, y_0)$. For the parabola $y^2=4ax$,the slope of the tangent $m_1$ at $(x_0, y_0)$ is given by $2y y' = 4a \Rightarrow m_1 = \frac{2a}{y_0}$.For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,the slope of the tangent $m_2$ at $(x_0, y_0)$ is given by $\frac{2x}{a^2} + \frac{2y y'}{b^2} = 0 \Rightarrow m_2 = -\frac{b^2 x_0}{a^2 y_0}$.Since they intersect at right angles,$m_1 m_2 = -1$,so $\left(\frac{2a}{y_0}\right) \left(-\frac{b^2 x_0}{a^2 y_0}\right) = -1 \Rightarrow \frac{2b^2 x_0}{a y_0^2} = 1$.Substituting $y_0^2 = 4ax_0$,we get $\frac{2b^2 x_0}{a(4ax_0)} = 1$ $\Rightarrow \frac{b^2}{2a^2} = 1$ $\Rightarrow b^2 = 2a^2$.The eccentricity $e$ of the ellipse $(b>a)$ is given by $a^2 = b^2(1-e^2)$ $\Rightarrow a^2 = 2a^2(1-e^2)$ $\Rightarrow 1 = 2(1-e^2)$ $\Rightarrow 1 = 2 - 2e^2$ $\Rightarrow 2e^2 = 1$.

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=4ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $2e^2=$

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