# The eccentric angle in the first quadrant of a point on the ellipse x^2/10 +y^2/8=1 at a distance of 3 units from the centre of the ellipse is

Solution:

$\\ a=\sqrt{10};b=\sqrt{8}$

Let a point $\\ P(\sqrt{10}cos\theta ,\sqrt{8}sin\theta )$ .its distance from $(0,0)$ is $3$.

$\\ \Rightarrow 10cos^2\theta +8sin^2\theta =9\\ \\\Rightarrow 2cos^2\theta =1\\ \\\Rightarrow cos\theta =1/\sqrt{2}\\ \\\Rightarrow \theta =\pi /4$

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