Thank you for reporting, we will resolve it shortly
Q.
Let $a$ and $b$ be any two numbers satisfying $\frac{1}{a^{2}}+\frac{1}{b^{2}} = \frac{1}{4}.$ Then, the foot of perpendicular from the origin on the variable line, $\frac{x}{a}+\frac{y}{b} = 1,$ lies on :
Equation of $\bot$
$\frac{h-0}{\frac{1}{a}}=\frac{k-0}{\frac{1}{b}}=\frac{1}{\frac{1}{a^{2}}+\frac{1}{b^{2}}}$
$\Rightarrow a=\frac{4}{h} \,......\left(i\right)$ and $b=\frac{4}{k} \,......\left(ii\right)$
to find locus of $\left(h, k\right)$ put
$\left(i\right)$ and $\left(ii\right)$ in $\frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{1}{4}$
i.e. $\frac{h^{2}}{16}+\frac{k^{2}}{16}=\frac{1}{4}$
$\Rightarrow $ locus is $\frac{x^{2}}{16}+\frac{y^{2}}{16}=\frac{1}{4}$
$\Rightarrow x^{2}+y^{2}=4$