frac{x^2}{a^2} + frac{y^2}{b^2} = 1 (b a) i | vedclass

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(D) Given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b > a$ and eccentricity $e = \frac{1}{\sqrt{2}}$.Since $e^2 = 1 - \frac{a^2}{b^2}$

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$(\frac{a}{2}, \frac{\sqrt{3}a}{\sqrt{2}})$

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(D) Given the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $b > a$ and eccentricity $e = \frac{1}{\sqrt{2}}$.Since $e^2 = 1 - \frac{a^2}{b^2}$,we have $\frac{1}{2} = 1 - \frac{a^2}{b^2}$,which implies $\frac{a^2}{b^2} = \frac{1}{2}$,so $b^2 = 2a^2$.The ellipse equation is $\frac{x^2}{a^2} + \frac{y^2}{2a^2} = 1$,or $2x^2 + y^2 = 2a^2$.For the parabola $y^2 = 4ax$,substituting $y^2$ into the ellipse equation: $2x^2 + 4ax - 2a^2 = 0$,or $x^2 + 2ax - a^2 = 0$.Solving for $x$: $x = \frac{-2a \pm \sqrt{4a^2 + 4a^2}}{2} = -a \pm a\sqrt{2}$. Since $x > 0$,$x = a(\sqrt{2} - 1)$.Then $y^2 = 4a^2(\sqrt{2} - 1)$,so $y = 2a\sqrt{\sqrt{2} - 1}$.The slopes $m_1$ (ellipse) and $m_2$ (parabola) at the intersection point $(x_0, y_0)$ are found by differentiation.For $2x^2 + y^2 = 2a^2$,$4x + 2y y' = 0 \implies m_1 = -\frac{2x_0}{y_0}$.For $y^2 = 4ax$,$2y y' = 4a \implies m_2 = \frac{2a}{y_0}$.$	an 	heta = |\frac{m_1 - m_2}{1 + m_1 m_2}| = |\frac{-2x_0 - 2a}{y_0(1 - \frac{4ax_0}{y_0^2})}| = |\frac{-2(x_0 + a)}{y_0(1 - \frac{4ax_0}{4ax_0})}|$ is undefined,meaning the curves intersect at $90^\circ$ or $	heta = \frac{\pi}{2}$.Thus,$	heta = \frac{\pi}{2}$. The point on the ellipse is determined by the parameter $\phi$ where $x = a \cos \phi, y = b \sin \phi = a\sqrt{2} \sin \phi$. For $	heta = \frac{\pi}{2}$,the coordinates are $(\frac{a}{2}, \frac{\sqrt{3}a}{\sqrt{2}})$.

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(b > a)$ is an ellipse with eccentricity $e = \frac{1}{\sqrt{2}}$. If the angle of intersection between the ellipse and the parabola $y^2 = 4ax$ is $\theta$,then the coordinates of the point corresponding to $\theta$ on the ellipse are:

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