Question

4) Find the eccentric angle of the point
(a) (4, 3) on the ellipse $\frac{x^2}{64}$+$\frac{y^2}{12}$ = 1 ,
(b) (x₁, y₁) on the ellipse $\frac{x^2}{6}+\frac{y^2}{2}$ = 1 whose distance from the centre is 2.

Answer 1

Dear student
(ii)
$$\begin{gathered} \mathrm{Let}\mathrm{\theta }\mathrm{be}\mathrm{the}\mathrm{eccentric}\mathrm{angle}\mathrm{of}\mathrm{the}\mathrm{point}\mathrm{P}. \\ \mathrm{Then}\mathrm{the}\mathrm{co}-\mathrm{ordinate}\mathrm{of}\mathrm{P}\mathrm{are}\left(\sqrt{6}\mathrm{cos\theta },\sqrt{2}\mathrm{sin\theta }\right) \\ \mathrm{The}\mathrm{centre}\mathrm{of}\mathrm{the}\mathrm{ellipse}\mathrm{is}\mathrm{at}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{origin}. \\ \mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that} \\ \mathrm{OP}=2\mathrm{where}\mathrm{O}\mathrm{is}\mathrm{the}\mathrm{origin}. \\ \Rightarrow \sqrt{6{\cos }^2\mathrm{\theta }+2{\sin }^2\mathrm{\theta }}=2 \\ \Rightarrow 6{\cos }^2\mathrm{\theta }+2{\sin }^2\mathrm{\theta }=4 \\ \Rightarrow 3{\cos }^2\mathrm{\theta }+{\sin }^2\mathrm{\theta }=2 \\ \Rightarrow {\sin }^2\mathrm{\theta }+{\cos }^2\mathrm{\theta }+2{\cos }^2\mathrm{\theta }=2 \\ \Rightarrow 1+2{\cos }^2\mathrm{\theta }=2 \\ \Rightarrow 2{\cos }^2\mathrm{\theta }=1 \\ \Rightarrow {\cos }^2\mathrm{\theta }=\frac{1}{2} \\ \Rightarrow \mathrm{cos\theta }=\pm \frac{1}{\sqrt{2}} \\ \Rightarrow \mathrm{\theta }=\pm \frac{\mathrm{\pi }}{4} \end{gathered}$$
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