The locus of the middle point of the intercep | vedclass

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(C) The given ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$. Here $a^2 = 2$ and $b^2 = 1$.Let the point of c

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The given ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$. Here $a^2 = 2$ and $b^2 = 1$.Let the point of contact be $R \equiv (\sqrt{2} \cos 	heta, \sin 	heta)$.The equation of the tangent at $R$ is $\frac{x \cos 	heta}{\sqrt{2}} + y \sin 	heta = 1$.The tangent intersects the $x$-axis at $A \equiv (\sqrt{2} \sec 	heta, 0)$ and the $y$-axis at $B \equiv (0, \csc 	heta)$.Let $Q(h, k)$ be the midpoint of $AB$. Then $h = \frac{\sqrt{2} \sec 	heta}{2} = \frac{\sec 	heta}{\sqrt{2}}$ and $k = \frac{\csc 	heta}{2}$.This implies $\cos 	heta = \frac{1}{h\sqrt{2}}$ and $\sin 	heta = \frac{1}{2k}$.Using $\cos^2 	heta + \sin^2 	heta = 1$,we get $\left(\frac{1}{h\sqrt{2}}ight)^2 + \left(\frac{1}{2k}ight)^2 = 1$.Thus,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

The locus of the middle point of the intercept of the tangents drawn to the ellipse $x^2 + 2y^2 = 2$ between the coordinate axes is:

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