Let a tangent to the curve 9x^2 + 16y^2 = 144 | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Any point $P$ on the ellipse can be represented as $(4 \cos 	heta, 3 \sin

Vedclass · Accepted Answer

$7$

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(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{9} = 1$. Any point $P$ on the ellipse can be represented as $(4 \cos 	heta, 3 \sin 	heta)$. The equation of the tangent at $P$ is $\frac{x \cos 	heta}{4} + \frac{y \sin 	heta}{3} = 1$. The coordinates of $A$ (where $y=0$) are $(4 \sec 	heta, 0)$ and $B$ (where $x=0$) are $(0, 3 \operatorname{cosec} 	heta)$. The length $L$ of the segment $AB$ is given by $L^2 = 16 \sec^2 	heta + 9 \operatorname{cosec}^2 	heta$. Using the identity $\sec^2 	heta = 1 + 	an^2 	heta$ and $\operatorname{cosec}^2 	heta = 1 + \cot^2 	heta$,we get $L^2 = 16(1 + 	an^2 	heta) + 9(1 + \cot^2 	heta) = 25 + 16 	an^2 	heta + 9 \cot^2 	heta$. By the $AM$-$GM$ inequality,$16 	an^2 	heta + 9 \cot^2 	heta \geq 2 \sqrt{16 	an^2 	heta \cdot 9 \cot^2 	heta} = 2 \cdot 4 \cdot 3 = 24$. Thus,$L^2 \geq 25 + 24 = 49$,which implies $L \geq 7$. The minimum length of the line segment $AB$ is $7$.

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

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