The eccentric angle in the first quadrant of | vedclass

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(B) Let $P(\sqrt{10} \cos 	heta, \sqrt{8} \sin 	heta)$ be the point on the ellipse $\frac{x^{2}}{10}+\frac{y^{2}}{8}=1$.Given that the distance of $

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$\frac{\pi}{4}$

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(B) Let $P(\sqrt{10} \cos 	heta, \sqrt{8} \sin 	heta)$ be the point on the ellipse $\frac{x^{2}}{10}+\frac{y^{2}}{8}=1$.Given that the distance of $P$ from the centre $(0,0)$ is $3$ units.So,$OP^2 = 3^2 = 9$.$10 \cos^2 	heta + 8 \sin^2 	heta = 9$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we can write $9 = 9(\sin^2 	heta + \cos^2 	heta)$.$10 \cos^2 	heta + 8 \sin^2 	heta = 9 \sin^2 	heta + 9 \cos^2 	heta$.Rearranging the terms,we get $\cos^2 	heta = \sin^2 	heta$.$	an^2 	heta = 1$.Since the point is in the first quadrant,$	heta = \frac{\pi}{4}$.

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

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