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(C) Given ellipse is $\frac{x^2}{64}+\frac{y^2}{49}=1$. Here,$a^2=64 \Rightarrow a=8$ and $b^2=49 \Rightarrow b=7$. The equation of a tangent to the e

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$15$

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(C) Given ellipse is $\frac{x^2}{64}+\frac{y^2}{49}=1$. Here,$a^2=64 \Rightarrow a=8$ and $b^2=49 \Rightarrow b=7$. The equation of a tangent to the ellipse at any point $(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$. The intercepts on the coordinate axes are $x_0 = \frac{a}{\cos 	heta}$ and $y_0 = \frac{b}{\sin 	heta}$. The sum of the intercepts is $L = a \sec 	heta + b \csc 	heta$. To find the minimum value,we differentiate $L$ with respect to $	heta$ and set it to $0$: $\frac{dL}{d	heta} = a \sec 	heta 	an 	heta - b \csc 	heta \cot 	heta = 0$. This gives $	an^3 	heta = \frac{b}{a}$,so $	an 	heta = (b/a)^{1/3}$. The minimum sum of intercepts is $a+b = 8+7 = 15$.

The least intercept made by a tangent to the ellipse $\frac{x^2}{64}+\frac{y^2}{49}=1$ with the coordinate axes is

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