The locus of the mid-points of the points of | vedclass

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(B) The given line is $x \cos 	heta + y \sin 	heta = 1$. To find the intersection points with the coordinate axes,we set $y = 0$ and $x = 0$ respect

Vedclass · Accepted Answer

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

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(B) The given line is $x \cos 	heta + y \sin 	heta = 1$. To find the intersection points with the coordinate axes,we set $y = 0$ and $x = 0$ respectively. For $y = 0$,$x = \frac{1}{\cos 	heta}$,so point $A = (\frac{1}{\cos 	heta}, 0)$. For $x = 0$,$y = \frac{1}{\sin 	heta}$,so point $B = (0, \frac{1}{\sin 	heta})$. Let $(h, k)$ be the mid-point of $AB$. Then $h = \frac{1}{2 \cos 	heta}$ and $k = \frac{1}{2 \sin 	heta}$. This implies $\cos 	heta = \frac{1}{2h}$ and $\sin 	heta = \frac{1}{2k}$. Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we get $(\frac{1}{2h})^2 + (\frac{1}{2k})^2 = 1$. $\frac{1}{4h^2} + \frac{1}{4k^2} = 1$,which simplifies to $\frac{1}{h^2} + \frac{1}{k^2} = 4$. Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{x^2} + \frac{1}{y^2} = 4$.

The locus of the mid-points of the points of intersection of $x \cos \theta + y \sin \theta = 1$ with the coordinate axes is

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