Question

A right angle triangle ABC having right angle at C, CA = a, CB =b moves such that the angular points A and B slide along x-axis and y-axis respectively. The locus of C is

• A. $ax+by+1=0$
• B. $ax\pm by=0$
• C. $a{x}^2\pm 2by+y^2=0$
• D. none of these

Answer 1

The correct option is B $ax\pm by=0$

Let the point $C$ be $(h,K)$

and $A\equiv (l,0),B\equiv (0,m)$

Slope of $AC=\frac{k}{h-l}⟶\left(1\right)$

Slope of $BC=\frac{K-m}{h}⟶\left(2\right)$

Using distance formula,

$a=\sqrt{{(h-l)}^2+K^2}$

$\Rightarrow h-l=\sqrt{a^2-K^2}⟶\left(3\right)$

Similarly

$K-m=\sqrt{b^2-h^2}⟶\left(4\right)$

Slope of $AC=\frac{K}{\sqrt{a^2-K^2}}$

Slope of $BC=\frac{\sqrt{b^2-h^2}}{h}$

Since $AB$ is perpendicular to $BC$

$\therefore$ $\frac{K}{\sqrt{a^2-K^2}}.\frac{\sqrt{b^2-h^2}}{h}=-1$

Squaring both sides we get

$\frac{K^2}{a^2-K^2}=\frac{h^2}{b^2-h^2}$

$(b^2-h^2)K^2=h^2(a^2-K^2)$

$bK\pm ah=0$

Hence locus of point $C$ is

$ax\pm by=0$

Option B.