Question

A variable straight line draw through the point of intersection of the straight lines
$\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$
meets the coordination axes in $A$ and $B$. Show that the locus of the mid-point of $AB$ is the curve
$2xy(a+b)=ab(x+y)$.

Answer 1

the point of intersection of given lines is $(\frac{ab}{a+b},\frac{ab}{a+b})$. Any line through this point is
$(y-\frac{ab}{a+b})=m(x-\frac{ab}{a+b})$
Putting $y=0$ and than $x=0$ the points where it meet the axes are
$A(y-\frac{ab}{a+b}\frac{m-1}{m},0),B(0-\frac{ab}{a+b}(m-1))$
If $(h,k)$ be the mid-point of $AB$, then
$2h=\frac{ab}{a+b}\frac{m-1}{m},2k=\frac{ab}{a+b}(m-1)$
In order to find the locus of $(h,k)$ we have to eliminate the variable $m$.
$\therefore \frac{1}{2h},\frac{1}{2k}=\frac{a+b}{ab}[\frac{m}{m-1}-\frac{1}{m-1}]=\frac{a+b}{ab}$
Hence the locus is $ab(x+y)=2xy(a+b)$.