The equation to the locus of the middle point | vedclass

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Question

(A) The equation of any tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is given by $\frac{x \cos 	heta}{4} + \frac{y \sin 	heta}{3} = 1

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$9x^2 + 16y^2 = 4x^2y^2$

Vedclass · Answer

(A) The equation of any tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is given by $\frac{x \cos 	heta}{4} + \frac{y \sin 	heta}{3} = 1$.The tangent meets the coordinate axes at points $A\left(\frac{4}{\cos 	heta}, 0ight)$ and $B\left(0, \frac{3}{\sin 	heta}ight)$.Let the midpoint of $AB$ be $(h, k)$. Then,$h = \frac{2}{\cos 	heta} \implies \cos 	heta = \frac{2}{h}$$k = \frac{3}{2 \sin 	heta} \implies \sin 	heta = \frac{3}{2k}$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we get:$\left(\frac{2}{h}ight)^2 + \left(\frac{3}{2k}ight)^2 = 1$$\frac{4}{h^2} + \frac{9}{4k^2} = 1$Multiplying by $4h^2k^2$,we obtain:$16k^2 + 9h^2 = 4h^2k^2$.Replacing $(h, k)$ with $(x, y)$,the locus is $9x^2 + 16y^2 = 4x^2y^2$.

The equation to the locus of the middle point of the portion of the tangent to the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ included between the coordinate axes is:

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