The angle between ellipse $\frac{x^{2}}{4} + y^{2} = 1$ and circle $x^{2} + y^{2} = 2$ is $ heta$,

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The angle between ellipse $\frac{x^{2}}{4} + y^{2} = 1$ and circle $x^{2} + y^{2} = 2$ is $	heta$, then $tan	heta$ is equal to:

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Sol.(b) Equation of normal to ellipse at $P$ is$$\frac{4x}{2/\sqrt{3}} - \frac{y}{\sqrt{2}/\sqrt{3}} = 3$$$$\Rightarrow 2x - \frac{y}{\sqrt{2}} = \sqrt{3} $$⇒ slope of normal to ellipse at $P = m_{1} = 2\sqrt{2}$Slope of normal to circle at$$\begin{matrix} P = m_{2} = \frac{\frac{\sqrt{2}}{\sqrt{3}} - 0}{\frac{2}{\sqrt{3}} - 0} = \frac{1}{\sqrt{2}} \ tan	heta = \frac{2\sqrt{2} - \frac{1}{\sqrt{2}}}{1 + 2\sqrt{2}\frac{1}{\sqrt{2}}} = \frac{3}{3\sqrt{2}} = \frac{1}{\sqrt{2}} \end{matrix}$$

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