Question
If the inclination of a diameter $PP^{'}$ of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is $\theta$ and $\left( PP^{'} \right)^{2}$ is the A.M. of squares of the lengths of major and minor axes, then $tan\theta$ is equal to :
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Solution
Sol.(a) $r^{2}\left( \frac{\cos^{2}\theta}{a^{2}} + \frac{\sin^{2}\theta}{b^{2}} \right) = 1$
$$\begin{matrix} & \ 2(2r)^{2} = (2a)^{2} + (2b)^{2} \\ \Rightarrow \ & \ 2r^{2} = a^{2} + b^{2} \\ \Rightarrow \ & \ \left( a^{2} + b^{2} \right)\left( \frac{\cos^{2}\theta}{a^{2}} + \frac{\sin^{2}\theta}{b^{2}} \right) = 2 \\ \Rightarrow & \ \frac{b^{2}}{a^{2}}\cos^{2}\theta + \frac{a^{2}}{b^{2}}\sin^{2}\theta = 1 = \sin^{2}\theta + \cos^{2}\theta \\ \ & \\ \Rightarrow & \left( \frac{b^{2} - a^{2}}{a^{2}} \right)\cos^{2}\theta = \left( 1 - \frac{a^{2}}{b^{2}} \right)\sin^{2}\theta \\ \Rightarrow & tan\theta = \pm \frac{b}{a} \end{matrix}$$
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