Question

If the variable line $\frac{x}{\alpha }+\frac{y}{\beta }=1$($\alpha ,\beta$ are variables) moves in such a way that $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{1}{m^2}$, (m is constant, then find the locus of foot of perpendicular from the origin on the straight line is?

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

Answer 1

The correct option is A $x^2+y^2=m^2$

$\text{Solve:-}$

$\text{Given,}\frac{x}{\alpha }+\frac{y}{\beta }=1$

$\text{Let}(h,k)\text{be the foot of the perpendicular}$

$\text{from origin to Given line}$

$\Rightarrow \frac{0-h}{\frac{1}{\alpha }}=\frac{0-k}{\frac{1}{\beta }}=\frac{+(\frac{1}{\alpha }\times 0+\frac{1}{\beta }\times 0-1)}{\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}}$

by using $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=\frac{1}{m^2}$ we get,

$h=\frac{m^2}{\alpha },k=\frac{m^2}{\beta }$

$\Rightarrow h^2+k^2=\frac{m^4}{{\alpha }^2}+\frac{m^4}{{\beta }^2}$

$\Rightarrow h^2+k^2=m^4[\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}]$

$=m^4\times \frac{1}{m^2}=m^2$

$\Rightarrow x^2+y^2=m^2$ is the required

locus.