Let a circle of radius 4 and the ellipse 15x^ | vedclass

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(B) The equation of the ellipse is $15x^2 + 19y^2 = 285$,which can be written as $\frac{x^2}{19} + \frac{y^2}{15} = 1$.Here,$a^2 = 19$ and $b^2 = 15$.

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$\frac{\pi}{3}$

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(B) The equation of the ellipse is $15x^2 + 19y^2 = 285$,which can be written as $\frac{x^2}{19} + \frac{y^2}{15} = 1$.Here,$a^2 = 19$ and $b^2 = 15$.The equation of a tangent to the ellipse with slope $m$ is $y = mx \pm \sqrt{a^2m^2 + b^2}$,which is $y = mx \pm \sqrt{19m^2 + 15}$.This line is also a tangent to the circle $x^2 + y^2 = 4^2 = 16$,so its perpendicular distance from the center $(0,0)$ must be equal to the radius $4$.$\left| \frac{\pm \sqrt{19m^2 + 15}}{\sqrt{m^2 + 1}} ight| = 4$.Squaring both sides,we get $\frac{19m^2 + 15}{m^2 + 1} = 16$.$19m^2 + 15 = 16m^2 + 16$,which simplifies to $3m^2 = 1$,so $m = \pm \frac{1}{\sqrt{3}}$.The angle $	heta$ that the tangent makes with the $x$-axis is given by $	an 	heta = |m| = \frac{1}{\sqrt{3}}$,so $	heta = \frac{\pi}{6}$.The minor axis of the ellipse is the $y$-axis (since $b^2 The angle between the tangent and the $y$-axis is $\frac{\pi}{2} - 	heta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$.

Let a circle of radius $4$ and the ellipse $15x^2 + 19y^2 = 285$ be concentric. What is the angle that the common tangents make with the minor axis of the ellipse?

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