Question

A verible line drawn through the points of intersection of the lines $\frac{x}{a}+\frac{y}{b}=1,\frac{x}{b}+\frac{y}{a}=1$ meets the coordinates axes in $A$ and $B$.Then the locus of the mid point of $AB$ is

• A. $2xy(a+b)=ab(x+y)$
• B. $xy(a+b)=ab(x-y)$
• C. $xy(a+b)=ab(x+y)$
• D. $xy(a-b)=ab(x+y)$

Answer 1

The correct option is A $2xy(a+b)=ab(x+y)$

Equation of line $AB$ is

$(bx+ay-ab)+\lambda (ax+by-ab)=0$

when $x=0$

$ay-ab+\lambda (by-ab)=0$

when $y=0$ or, $y(a+\lambda b)-ab(\lambda +1)=0$

$(bx+0-ab)+\lambda (ax-ab)=0$ or, $y=\frac{ab(\lambda +1)}{a+\lambda b}$

or, $bx+\lambda ax-ab(1+\lambda )=0$ Midpoint $P(h,k)$ is given by

$\therefore x=\frac{ab(1+\lambda )}{b+a\lambda }$ $h=\frac{ab(1+\lambda )}{\frac{b+a\lambda }{2}}+0$

and $k=\frac{ab(\lambda +1}{\frac{a+\lambda b}{2}}+0$

or, $2h\frac{ab(1+\lambda )}{b+a\lambda }$ and $2k=\frac{ab(1+\lambda )}{1+\lambda b}$

Now, $\frac{1}{2h}+\frac{2}{k}=\frac{b+a\lambda }{ab(1+\lambda )}+\frac{a+\lambda b}{ab(1+\lambda )}$

$=\frac{(b+a)+\lambda (b+a)}{ab(1+\lambda )}=\frac{a+b}{ab}$

or, $\frac{1}{2}\left(\frac{h+k}{hk}\right)=\frac{a+b}{ab}\Rightarrow ab(h+k)=2hk(a+b)$

$\therefore$ Locus is: $ab(x+y)=2xy(a+b)$