If P(alpha, beta) is a point on the curve 9x^ | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{36} = 1$,where $a^2 = 16$ and $b^2 = 36$. So $a=4$ and $b=6$.Let the point $P$ be $(4

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$S=24$

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(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{36} = 1$,where $a^2 = 16$ and $b^2 = 36$. So $a=4$ and $b=6$.Let the point $P$ be $(4 \cos 	heta, 6 \sin 	heta)$ for $	heta \in (0, \pi/2)$.The equation of the tangent at $P$ is $\frac{x \cos 	heta}{4} + \frac{y \sin 	heta}{6} = 1$.The intercepts on the axes are $x_0 = \frac{4}{\cos 	heta}$ and $y_0 = \frac{6}{\sin 	heta}$.The area of the triangle formed by the tangent and the coordinate axes is $A = \frac{1}{2} |x_0 y_0| = \frac{1}{2} \cdot \frac{4}{\cos 	heta} \cdot \frac{6}{\sin 	heta} = \frac{12}{\sin 	heta \cos 	heta} = \frac{24}{\sin(2	heta)}$.To minimize the area $A$,we maximize $\sin(2	heta)$. The maximum value of $\sin(2	heta)$ is $1$ at $2	heta = \pi/2$,i.e.,$	heta = \pi/4$.Thus,the minimum area $S = \frac{24}{1} = 24$.

If $P(\alpha, \beta)$ is a point on the curve $9x^2 + 4y^2 = 144$ in the first quadrant and the minimum area of the triangle formed by the tangent of the curve at $P$ with the coordinate axes is $S$,then

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