Question

If a variable straight line is drawn through the point of intersection of $\frac{x}{a}+\frac{y}{b}=1$ and $\frac{x}{b}+\frac{y}{a}=1$ meets the co-ordinate axes at $A$ and $B$, then the locus of the mid-point of line segment $AB$ is

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

Answer 1

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

Given lines
$L_1:\frac{x}{a}+\frac{y}{b}-1=0$ and $L_2:\frac{x}{b}+\frac{y}{a}-1=0$
Line passing through the point of intersection of both the lines is
$L_1+\lambda L_2=0\Rightarrow (\frac{x}{a}+\frac{y}{b}-1)+\lambda (\frac{x}{b}+\frac{y}{a}-1)=0\Rightarrow (bx+ay-ab)+\lambda (ax+by-ab)=0\Rightarrow (b+\lambda a)x+(a+\lambda b)y-ab(1+\lambda )=0$

Finding $A$ and $B$,
$A=(\frac{ab(1+\lambda )}{b+a\lambda },0)B=(0,\frac{ab(1+\lambda )}{a+b\lambda })$

We know that $(h,k)$ is the mid-point of $AB$, we get
$(h,k)=(\frac{ab(1+\lambda )}{2(b+a\lambda )},\frac{ab(1+\lambda )}{2(a+b\lambda )})\Rightarrow \frac{1}{2h}+\frac{1}{2k}=\frac{a+b+a\lambda +b\lambda }{ab(1+\lambda )}\Rightarrow \frac{1}{2h}+\frac{1}{2k}=\frac{a(1+\lambda )+b(1+\lambda )}{ab(1+\lambda )}\Rightarrow \frac{1}{2h}+\frac{1}{2k}=\frac{a+b}{ab}\Rightarrow \frac{h+k}{hk}=\frac{2(a+b)}{ab}$

Hence, the required locus is
$(x+y)ab=2xy(a+b)$