Question

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 :

• A. a hyperbola with each semi-axis = $\sqrt{2}$
• B. hyperbola with each semi-axis = 2
• C. a circle of radius = 2
• D. a circle of radius = $\sqrt{2}$

Answer 1

Answer: (C)

Equation of $\perp$
$\frac{h-0}{\frac{1}{a}}=\frac{k-0}{\frac{1}{b}}=\frac{1}{\frac{1}{a^2}+\frac{1}{b^2}}$
$\Rightarrow a=\frac{4}{h}......\left(i\right)$ 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 $\frac{x^2}{16}+\frac{y^2}{16}=\frac{1}{4}$
$\Rightarrow x^2+y^2=4$