Question

A variable line, drawn 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 coordinate 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

$\frac{x}{a}+\frac{y}{b}=1$

$\Rightarrow bx+ay-ab=0$ ...(i)

$\frac{x}{b}+\frac{y}{a}=1$

$\Rightarrow ax+by-ab=0$ ...(ii)

(i) $\times a$ $abx+a^{2}y-a^{2}b=0$

(ii) $\times b$ $\underset{–}{abx+b^{2}y-a{b}^2=0}$

$(a^2-b^2)y=a^{2}b-a{b}^2$

$\Rightarrow (a-b)(a+b)y=ab(a-b)$

$\Rightarrow y=\frac{ab}{a+b}$

$\therefore x=\frac{ab}{a+b}$

$y-\frac{ab}{a+b}=m(x-\frac{ab}{a+b})$

$\therefore A(\frac{ab}{a+b}\frac{(m-1)}{m},0),B(0,\frac{ab}{a+b}(1-m\left)\right)$

$\therefore A(\frac{ab}{2(a+b)}x\frac{(m-1)}{m},\frac{ab}{2(a+b)}x(1-m\left)\right)$

$x=\frac{ab(m-1)}{2(a+b)},y=\frac{ab(1-m)}{2(a+b)}$

Equation ${}^{\prime }{m}^{\prime }$

$2xy(a+b)=ab(x+y)$ (proved)