Question

A variable line makes intercepts on the co-ordinate axes, the sum of whose squares is constant and equal to $k$ , the locus of the foot of the perpendicular from the origin to this line is

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

Answer 1

The correct option is C ${(x{}^2+y^2)}^2(\frac{1}{x^2}+\frac{1}{y^2})=k$

Let $P(a,b)$ be the foot of perpendicular of the line so its slope must be $-\frac{a}{b}$
Its equation should be
$y-b=-\frac{a}{b}(x-a)\Rightarrow ax+by=a^2+b^2\Rightarrow \frac{x}{\frac{a^2+b^2}{a}}+\frac{y}{\frac{a^2+b^2}{b}}=1\Rightarrow {\left(\frac{a^2+b^2}{a}\right)}^2+{\left(\frac{a^2+b^2}{b}\right)}^2=k$
Therefore locus of $(a,b)$ is
${(x^2+y^2)}^2(\frac{1}{x^2}+\frac{1}{y^2})=k$