Question

A straight line $L_1:\frac{x}{a}+\frac{y}{b}=1$ intersects the $x$-axis and $y$-axis at $P$ and $Q$ respectively and a straight line $L_2$ perpendicular to $L_1$ cuts the $x$-axis and $y$-axis at $R$ and $S$ respectively. The locus of the point of intersection of the lines $PS$ and $QR$ is

• A. $x^2+y^2+ax-by=0$
• B. $x^2+y^2-ax-by=0$
• C. $x^2-y^2-ax-by=0$
• D. $x^2+y^2+ax+by=0$

Answer 1

The correct option is B $x^2+y^2-ax-by=0$

Given equation of the line $L_1$ be
$\frac{x}{a}+\frac{y}{b}=1\cdots \left(i\right)$
It cuts the coordinates axes at $P(a,0)$ and $Q(0,b)$.
The equation of line $L_2$ is
$\frac{x}{b}-\frac{y}{a}=\lambda$, where $\lambda$ is parameter
$\Rightarrow \frac{x}{b\lambda }-\frac{y}{a\lambda }=1$

It cuts the coordinate axes at
$R(b\lambda ,0)$ and $S(0,-a\lambda )$
Equation of line $PS$ and $QR$ are
$\frac{x}{a}-\frac{y}{a\lambda }=1$ and $\frac{x}{b\lambda }+\frac{y}{b}=1$

Let $T(h,k)$ be the point of intersection of $PS$ and $QR$
Then,
$\frac{h}{a}-\frac{k}{a\lambda }=1\text{and}\frac{h}{b\lambda }+\frac{k}{b}=1\Rightarrow \frac{h}{a}-1=\frac{k}{a\lambda }\text{and}\frac{h}{b\lambda }=1-\frac{k}{b}\Rightarrow \frac{h-a}{k}=\frac{1}{\lambda }\text{and}\frac{b-k}{h}=\frac{1}{\lambda }\Rightarrow \frac{h-a}{k}=\frac{b-k}{h}\Rightarrow h^2+k^2-ah-bk=0$

Hence, the locus of $T(h,k)$ is
$x^2+y^2-ax-by=0$