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(A) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$, where $a^2 = 25$ and $b^2 = 9$. Thus, $a = 5$ and $b = 3$.Any point on the el

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(A) The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$, where $a^2 = 25$ and $b^2 = 9$. Thus, $a = 5$ and $b = 3$.Any point on the ellipse can be represented as $(a \cos 	heta, b \sin 	heta) = (5 \cos 	heta, 3 \sin 	heta)$.The distance of this point from the center $(0, 0)$ is given as $5$.Using the distance formula: $\sqrt{(5 \cos 	heta - 0)^2 + (3 \sin 	heta - 0)^2} = 5$.Squaring both sides: $25 \cos^2 	heta + 9 \sin^2 	heta = 25$.Substitute $\cos^2 	heta = 1 - \sin^2 	heta$: $25(1 - \sin^2 	heta) + 9 \sin^2 	heta = 25$.$25 - 25 \sin^2 	heta + 9 \sin^2 	heta = 25$.$-16 \sin^2 	heta = 0$, which implies $\sin 	heta = 0$.Therefore, $	heta = 0$ or $\pi$.

If $\theta$ is the eccentric angle of a point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ such that the distance of the point from the center is $5$, then $\theta = ......$

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