Question

$p$ is the length of the perpendicular drawn from the origin upon a straight line then the locus of mid point of the portion of the line intercepted between the coordinate axes is

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

Answer 1

Answer: (C)

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

Let $y=mx+c$ line

distance from $(0,0)$ to $y=mx+c$ is $p$

$p=\left|\frac{c}{\sqrt{1+m^2}}\right|$

For $x$ intercept $y=0$

Therefore $x$ intercept : $x=\frac{-c}{m}$

For $y$ intercept $x=0$

$y$ intercept: $y=c$

$M$ is a mid-point of intercept made by line $y-mx-c=0$

$\therefore M\equiv (\frac{-c}{2m},\frac{c}{2})$

So, $x=\frac{-c}{2m},y=\frac{c}{2}$

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

$=\frac{4}{c^2}(1+m^2)$

$=\frac{4}{c^2}\times \frac{c^2}{p^2}$

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