Equation of given ellipse is
$\frac{x^2}{64}+\frac{y^2}{4}=1$
Now, let a parametric point on ellipse (i)
$P(8 \cos \theta, 2 \sin \theta)$
So, equation of tangent to ellipse (i) at point $P$ is
$\Rightarrow \quad \begin{aligned} & \frac{x \cos \theta}{8}+\frac{y \sin \theta}{2}=1 \\ & \frac{x}{\cos \theta}+\frac{y}{\frac{2}{\sin \theta}}=1\end{aligned}$
Now, length of the intercept between the coordinate axes made by the tangent (ii) is
$\begin{aligned} l & =\sqrt{\frac{64}{\cos ^2 \theta}+\frac{4}{\sin ^2 \theta}}=\sqrt{64 \sec ^2 \theta+4 \operatorname{cosec}^2 \theta} \\ & =\sqrt{68+64 \tan ^2 \theta+4 \cot ^2 \theta}\end{aligned}$
By AM-GM method, we have
$\begin{aligned} & \frac{64 \tan ^2 \theta+4 \cot ^2 \theta}{2} \geq \sqrt{64 \times 4} \\ & \Rightarrow 64 \tan ^2 \theta+4 \cot ^2 \theta \geq 32\end{aligned}$
So, the minimum value of $l$ is $\sqrt{68+32}=10$ Hence, option (a) is correct.