The eccentric angle of the point P(-6, 2) on | vedclass

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(C) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,any point $P$ on it can be represented as $(a \cos 	heta, b \sin 	heta)$,where $	heta$ i

Vedclass · Accepted Answer

$150$

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(C) For an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,any point $P$ on it can be represented as $(a \cos 	heta, b \sin 	heta)$,where $	heta$ is the eccentric angle.Given the equation $\frac{x^2}{48} + \frac{y^2}{16} = 1$,we have $a^2 = 48$ and $b^2 = 16$.Thus,$a = \sqrt{48} = 4\sqrt{3}$ and $b = \sqrt{16} = 4$.The point $P$ is $(-6, 2)$.Equating the coordinates: $a \cos 	heta = -6 \implies 4\sqrt{3} \cos 	heta = -6 \implies \cos 	heta = \frac{-6}{4\sqrt{3}} = \frac{-3}{2\sqrt{3}} = -\frac{\sqrt{3}}{2}$.Also,$b \sin 	heta = 2 \implies 4 \sin 	heta = 2 \implies \sin 	heta = \frac{2}{4} = \frac{1}{2}$.Since $\cos 	heta = -\frac{\sqrt{3}}{2}$ and $\sin 	heta = \frac{1}{2}$,the angle $	heta$ lies in the second quadrant.Therefore,$	heta = 180^{\circ} - 30^{\circ} = 150^{\circ}$.

The eccentric angle of the point $P(-6, 2)$ on the ellipse $\frac{x^2}{48} + \frac{y^2}{16} = 1$ is: (in $^{\circ}$)

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