If tangents are drawn from the point (2 + 13 | vedclass

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(C) The given ellipse is $\frac{(x-2)^2}{25} + \frac{(y-3)^2}{144} = 1$. Comparing this with the standard form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b

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$\frac{\pi}{2}$

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(C) The given ellipse is $\frac{(x-2)^2}{25} + \frac{(y-3)^2}{144} = 1$. Comparing this with the standard form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$,we have $a^2 = 25$ and $b^2 = 144$. The director circle of an ellipse is the locus of the point of intersection of perpendicular tangents to the ellipse,given by $(x-h)^2 + (y-k)^2 = a^2 + b^2$. Substituting the values,the equation of the director circle is $(x-2)^2 + (y-3)^2 = 25 + 144 = 169 = 13^2$. The given point $(2 + 13 \cos 	heta, 3 + 13 \sin 	heta)$ satisfies the equation $(x-2)^2 + (y-3)^2 = 13^2$,which means the point lies on the director circle. Since the point lies on the director circle,the tangents drawn from this point to the ellipse are perpendicular to each other. Therefore,the angle between the tangents is $\frac{\pi}{2}$.

If tangents are drawn from the point $(2 + 13 \cos \theta, 3 + 13 \sin \theta)$ to the ellipse $\frac{(x-2)^2}{25} + \frac{(y-3)^2}{144} = 1$,then the angle between them is:

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