Question

In a right angled isosceles $△ABC$, where $A$ is at origin and side lengths $AB$ and $AC$ are equal to $a$ units. If point $B$ and $C$ are produced to $P$ and $Q$ respectively such that $BP\cdot CQ=A{B}^2$, then the locus of the midpoint of $PQ$ is

• A. $\frac{1}{x}+\frac{1}{y}=\frac{4}{a}$
• B. $\frac{1}{x}+\frac{1}{y}=\frac{1}{2a}$
• C. $\frac{1}{x}+\frac{1}{y}=\frac{1}{a}$
• D. $\frac{1}{x}+\frac{1}{y}=\frac{2}{a}$

Answer 1

The correct option is D $\frac{1}{x}+\frac{1}{y}=\frac{2}{a}$

Assuming $A$ as the origin, $B=(a,0)$ and $C=(0,a)$ as shown in the below figure,
Let the coordinates of points $P=(h,0)$ and $Q=(0,k)$
Midpoint of $PQ$ is
$M=(\alpha ,\beta )=(\frac{h}{2},\frac{k}{2})$

Given, $BP.CQ=A{B}^2$
$\Rightarrow (h-a)(k-a)=a^2\Rightarrow hk-ak-ah+a^2=a^2\Rightarrow ak+ha=hk\Rightarrow \frac{1}{h}+\frac{1}{k}=\frac{1}{a}\Rightarrow \frac{1}{\alpha }+\frac{1}{\beta }=\frac{2}{a}(\because h=2\alpha ,k=2\beta )$

Hence, the locus of the midpoint is
$\frac{1}{x}+\frac{1}{y}=\frac{2}{a}$