Find the locus of the midpoint of the portion | vedclass

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Question

(A) The given ellipse is $x^{2} + 2y^{2} = 2$,which can be written as $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$.Any point on the ellipse can be represen

Vedclass · Accepted Answer

$\frac{1}{2x^{2}} + \frac{1}{4y^{2}} = 1$

Vedclass · Answer

(A) The given ellipse is $x^{2} + 2y^{2} = 2$,which can be written as $\frac{x^{2}}{2} + \frac{y^{2}}{1} = 1$.Any point on the ellipse can be represented as $(\sqrt{2} \cos 	heta, \sin 	heta)$.The equation of the tangent at this point is $\frac{x \cos 	heta}{\sqrt{2}} + \frac{y \sin 	heta}{1} = 1$.The tangent intersects the $x$-axis at $A(\sqrt{2} \sec 	heta, 0)$ and the $y$-axis at $B(0, \csc 	heta)$.Let the midpoint of the segment $AB$ be $Q(h, k)$.Then $h = \frac{\sqrt{2} \sec 	heta + 0}{2} = \frac{\sec 	heta}{\sqrt{2}}$ and $k = \frac{0 + \csc 	heta}{2} = \frac{\csc 	heta}{2}$.From these,we have $\sec 	heta = h\sqrt{2} \Rightarrow \cos 	heta = \frac{1}{h\sqrt{2}}$ and $\csc 	heta = 2k \Rightarrow \sin 	heta = \frac{1}{2k}$.Using the identity $\cos^{2} 	heta + \sin^{2} 	heta = 1$,we get $\left(\frac{1}{h\sqrt{2}}ight)^{2} + \left(\frac{1}{2k}ight)^{2} = 1$.Thus,$\frac{1}{2h^{2}} + \frac{1}{4k^{2}} = 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^{2}} + \frac{1}{4y^{2}} = 1$.

Find the locus of the midpoint of the portion of any tangent to the ellipse $x^{2} + 2y^{2} = 2$ intercepted between the axes.

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