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(C) The equation of the tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{(2a)^{2}}=1$ at point $(b \cos 	heta, 2a \sin 	heta)$ is given by $

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$2$

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(C) The equation of the tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{(2a)^{2}}=1$ at point $(b \cos 	heta, 2a \sin 	heta)$ is given by $\frac{x \cos 	heta}{b} + \frac{y \sin 	heta}{2a} = 1$.The $x$-intercept is $A = (\frac{b}{\cos 	heta}, 0)$ and the $y$-intercept is $B = (0, \frac{2a}{\sin 	heta})$.The area of the triangle $\Delta OAB$ is given by $	ext{Area} = \frac{1}{2} 	imes |x_{	ext{intercept}}| 	imes |y_{	ext{intercept}}| = \frac{1}{2} 	imes \frac{b}{\cos 	heta} 	imes \frac{2a}{\sin 	heta} = \frac{ab}{\sin 	heta \cos 	heta} = \frac{2ab}{\sin 2	heta}$.Since the minimum value of $\sin 2	heta$ is $1$ (for $	heta = \frac{\pi}{4}$),the minimum area is $2ab$.Given that the minimum area is $kab$,we have $kab = 2ab$,which implies $k = 2$.

If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{4 a^{2}}=1$ and the coordinate axes is $kab$,then $k$ is equal to ..... .

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