The locus of the mid-point of the segment int | vedclass

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(B) The given line is $x \cos \alpha + y \sin \alpha = p$.To find the $x$-intercept,set $y = 0$: $x \cos \alpha = p \Rightarrow x = \frac{p}{\cos \alp

Vedclass · Accepted Answer

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

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(B) The given line is $x \cos \alpha + y \sin \alpha = p$.To find the $x$-intercept,set $y = 0$: $x \cos \alpha = p \Rightarrow x = \frac{p}{\cos \alpha}$. So,point $A = (\frac{p}{\cos \alpha}, 0)$.To find the $y$-intercept,set $x = 0$: $y \sin \alpha = p \Rightarrow y = \frac{p}{\sin \alpha}$. So,point $B = (0, \frac{p}{\sin \alpha})$.Let $(h, k)$ be the mid-point of the segment $AB$.Then $h = \frac{p}{2 \cos \alpha}$ and $k = \frac{p}{2 \sin \alpha}$.This implies $\cos \alpha = \frac{p}{2h}$ and $\sin \alpha = \frac{p}{2k}$.Using the identity $\sin^2 \alpha + \cos^2 \alpha = 1$,we get $(\frac{p}{2k})^2 + (\frac{p}{2h})^2 = 1$.$\frac{p^2}{4k^2} + \frac{p^2}{4h^2} = 1 \Rightarrow \frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2}$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

The locus of the mid-point of the segment intercepted between the axes by the variable line $x \cos \alpha + y \sin \alpha = p$,where $p$ is a constant,is

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