The eccentric angle of the point (2, sqrt{3}) | vedclass

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(C) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$. Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get

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$\frac{\pi}{3}$

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(C) The given equation of the ellipse is $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$. Comparing this with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$,we get $a^{2}=16$ and $b^{2}=4$,so $a=4$ and $b=2$. Let $	heta$ be the eccentric angle of the point $P(2, \sqrt{3})$. The parametric coordinates of any point on the ellipse are given by $x = a \cos 	heta$ and $y = b \sin 	heta$. Substituting the values,we have $x = 4 \cos 	heta$ and $y = 2 \sin 	heta$. Given the point $(2, \sqrt{3})$,we equate: $2 = 4 \cos 	heta \Rightarrow \cos 	heta = \frac{2}{4} = \frac{1}{2}$ $\sqrt{3} = 2 \sin 	heta \Rightarrow \sin 	heta = \frac{\sqrt{3}}{2}$ Since both $\sin 	heta = \frac{\sqrt{3}}{2}$ and $\cos 	heta = \frac{1}{2}$ are satisfied at $	heta = \frac{\pi}{3}$,the eccentric angle is $\frac{\pi}{3}$.

The eccentric angle of the point $(2, \sqrt{3})$ lying on $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$ is

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