Question

Locus of midpoint of the portion of $x\cos a+y\sin a=p$ intercepted between the coordinate axes, where $p$ is a constant, is

• A. $x^2+y^2=\frac{4}{p^2}$
• B. $x^2+y^2=4p^2$
• C. $\frac{1}{x^2}+\frac{1}{y^2}=\frac{2}{p^2}$
• D. $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$

Answer 1

The correct option is D $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$

Given,

$x\cos \alpha +y\sin \alpha =p$

Let $P(h,k)$ be the mid point of above line.

when $x=0$

$y\sin \alpha =p$

$y=\frac{p}{\sin \alpha }$

when $y=0$

$x\cos \alpha =p$

$x=\frac{p}{\cos \alpha }$

Therefore the line meets the coordinate axes at $A(\frac{p}{\cos \alpha },0),B(0,\frac{p}{\sin \alpha })$

Clearly the mid point of the portion $AB=P$

$(\frac{\frac{p}{\cos \alpha }+0}{2},\frac{0+\frac{p}{\sin \alpha }}{2})=(h,k)$

$\Rightarrow \cos \alpha =\frac{p}{2h}$

$\Rightarrow \sin \alpha =\frac{p}{2k}$

squaring and adding the above equations, we get,

${\cos }^2\alpha +{\sin }^2\alpha =\frac{p^2}{4h^2}+\frac{p^2}{4k^2}$

$1=\frac{p^2}{4}(\frac{1}{h^2}+\frac{1}{k^2})$

$\frac{4}{p^2}=\frac{1}{h^2}+\frac{1}{k^2}$

The locus of P is $\frac{1}{x^2}+\frac{1}{y^2}=\frac{4}{p^2}$