Let a tangent to the curve 9x^2+16y^2=144 intersect the coordinate axes at points A and B.

Question

Let a tangent to the curve $9x^2+16y^2=144$ intersect the coordinate axes at points $A$ and $B$. Then the minimum length of the line segment $AB$ is:

Answer: 7

Accepted Answer

The ellipse is $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$. Tangent at $(4\cos\theta,3\sin\theta)$: $\dfrac{x\cos\theta}{4}+\dfrac{y\sin\theta}{3}=1$, meeting the axes at $A=(4\sec\theta,0)$, $B=(0,3\csc\theta)$.

$$AB^2=16\sec^2\theta+9\csc^2\theta=25+16\tan^2\theta+9\cot^2\theta\ge25+2\sqrt{16\cdot9}=49$$

by AM--GM, so $AB\ge7$, with equality attainable.

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