The minimum length of the intercept between t | vedclass

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Question

(A) The equation of the given ellipse is $\frac{x^2}{64}+\frac{y^2}{4}=1$. Let a parametric point on the ellipse be $P(8 \cos 	heta, 2 \sin 	heta)$.

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(A) The equation of the given ellipse is $\frac{x^2}{64}+\frac{y^2}{4}=1$. Let a parametric point on the ellipse be $P(8 \cos 	heta, 2 \sin 	heta)$. The equation of the tangent to the ellipse at point $P$ is $\frac{x \cos 	heta}{8}+\frac{y \sin 	heta}{2}=1$. This can be rewritten as $\frac{x}{8/\cos 	heta} + \frac{y}{2/\sin 	heta} = 1$. The length of the intercept $l$ between the coordinate axes is $l = \sqrt{(\frac{8}{\cos 	heta})^2 + (\frac{2}{\sin 	heta})^2} = \sqrt{64 \sec^2 	heta + 4 \operatorname{cosec}^2 	heta}$. Using $\sec^2 	heta = 1 + 	an^2 	heta$ and $\operatorname{cosec}^2 	heta = 1 + \cot^2 	heta$,we get $l = \sqrt{64(1 + 	an^2 	heta) + 4(1 + \cot^2 	heta)} = \sqrt{68 + 64 	an^2 	heta + 4 \cot^2 	heta}$. By the $AM-GM$ inequality,$64 	an^2 	heta + 4 \cot^2 	heta \geq 2 \sqrt{64 	an^2 	heta \cdot 4 \cot^2 	heta} = 2 \sqrt{256} = 32$. Thus,the minimum value of $l$ is $\sqrt{68 + 32} = \sqrt{100} = 10$. Therefore,option $A$ is correct.

The minimum length of the intercept between the coordinate axes made by a tangent of the ellipse $\frac{x^2}{64}+\frac{y^2}{4}=1$ is

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