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(D) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Here,$a^2 = 16$ and $b^2 = 9$.The director circle of an ellipse $\frac{x^2}{a

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$\frac{\pi}{2}$

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(D) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Here,$a^2 = 16$ and $b^2 = 9$.The director circle of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $x^2 + y^2 = a^2 + b^2$.Substituting the values,the director circle is $x^2 + y^2 = 16 + 9 = 25$.Now,check if point $P(x, y)$ lies on the director circle:$x = 3 \sin 	heta + 4 \cos 	heta$$y = 3 \cos 	heta - 4 \sin 	heta$$x^2 + y^2 = (3 \sin 	heta + 4 \cos 	heta)^2 + (3 \cos 	heta - 4 \sin 	heta)^2$$x^2 + y^2 = 9 \sin^2 	heta + 16 \cos^2 	heta + 24 \sin 	heta \cos 	heta + 9 \cos^2 	heta + 16 \sin^2 	heta - 24 \sin 	heta \cos 	heta$$x^2 + y^2 = 9(\sin^2 	heta + \cos^2 	heta) + 16(\cos^2 	heta + \sin^2 	heta) = 9(1) + 16(1) = 25$.Since $x^2 + y^2 = 25$,the point $P$ lies on the director circle.The angle between tangents drawn from any point on the director circle to the ellipse is always $\frac{\pi}{2}$.

If tangents are drawn from point $P(3 \sin \theta + 4 \cos \theta, 3 \cos \theta - 4 \sin \theta)$ where $\theta = \frac{\pi}{8}$ to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$,then the angle between the tangents is:

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