Find the locus of the midpoint of the portion | vedclass

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Any point on the ellipse is $(a \cos 	heta, b \sin 	heta)$.The equati

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$\frac{a^2}{x^2} + \frac{b^2}{y^2} = 4$

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Any point on the ellipse is $(a \cos 	heta, b \sin 	heta)$.The equation of the tangent at $(a \cos 	heta, b \sin 	heta)$ is $\frac{x \cos 	heta}{a} + \frac{y \sin 	heta}{b} = 1$.The tangent intersects the $x$-axis at $A(\frac{a}{\cos 	heta}, 0)$ and the $y$-axis at $B(0, \frac{b}{\sin 	heta})$.Let $(h, k)$ be the midpoint of $AB$.Then $h = \frac{a}{2 \cos 	heta} \implies \cos 	heta = \frac{a}{2h}$ and $k = \frac{b}{2 \sin 	heta} \implies \sin 	heta = \frac{b}{2k}$.Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we get $(\frac{a}{2h})^2 + (\frac{b}{2k})^2 = 1$.This simplifies to $\frac{a^2}{4h^2} + \frac{b^2}{4k^2} = 1$,or $\frac{a^2}{h^2} + \frac{b^2}{k^2} = 4$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{a^2}{x^2} + \frac{b^2}{y^2} = 4$.

Find the locus of the midpoint of the portion of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ intercepted between the axes.

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