The number of tangents to the circle x^2 + y^ | vedclass

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(A) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here $a^2 = 9$ and $b^2 = 4$.The equation of a normal to the ellipse at point

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(A) The equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{4} = 1$. Here $a^2 = 9$ and $b^2 = 4$.The equation of a normal to the ellipse at point $(3 \cos 	heta, 2 \sin 	heta)$ is given by $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$,which becomes $\frac{3x}{\cos 	heta} - \frac{2y}{\sin 	heta} = 9 - 4 = 5$.This can be rewritten as $3x \sec 	heta - 2y \csc 	heta = 5$.For this line to be a tangent to the circle $x^2 + y^2 = 3$,the perpendicular distance from the center $(0, 0)$ to the line must equal the radius $\sqrt{3}$.The distance $d = \frac{|-5|}{\sqrt{(3 \sec 	heta)^2 + (-2 \csc 	heta)^2}} = \frac{5}{\sqrt{9 \sec^2 	heta + 4 \csc^2 	heta}}$.Setting $d = \sqrt{3}$,we get $\sqrt{9 \sec^2 	heta + 4 \csc^2 	heta} = \frac{5}{\sqrt{3}}$,so $9 \sec^2 	heta + 4 \csc^2 	heta = \frac{25}{3} \approx 8.33$.Using the inequality $(m \sec^2 	heta + n \csc^2 	heta) \ge (\sqrt{m} + \sqrt{n})^2$,the minimum value of $9 \sec^2 	heta + 4 \csc^2 	heta$ is $(\sqrt{9} + \sqrt{4})^2 = (3 + 2)^2 = 25$.Since the minimum value $25$ is greater than $\frac{25}{3}$,there is no real value of $	heta$ that satisfies the condition.Therefore,the number of such tangents is $0$.

The number of tangents to the circle $x^2 + y^2 = 3$ that are normal to the ellipse $4x^2 + 9y^2 = 36$ is

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