Question

If a variable line drawn through the point of intersection of straight lines $\frac{x}{\alpha }+\frac{y}{\beta }=1$ and $\frac{x}{\beta }+\frac{y}{\alpha }=1$ meets the coordinate axes in $A$ and $B$, then the locus of the mid-point of $AB$ is:

• A. $\alpha \beta (x+y)=xy(\alpha +\beta )$
• B. $\alpha \beta (x+y)=2(\alpha +\beta )xy$
• C. $(\alpha +\beta )(x+y)=2\alpha \beta xy$
• D. None of these

Answer 1

The correct option is B $\alpha \beta (x+y)=2(\alpha +\beta )xy$

The equation of a line passing through the intersection of straight lines $\frac{x}{a}+\frac{y}{\beta }=1$ and $\frac{x}{\beta }+\frac{y}{a}=1$ is:

$(\frac{x}{\alpha }+\frac{y}{\beta }-1)+\lambda (\frac{x}{\beta }+\frac{y}{\alpha }-1)=0$

$\Rightarrow x(\frac{1}{\alpha }+\frac{\lambda }{\beta })+y(\frac{1}{\beta }+\frac{\lambda }{\alpha })-\lambda -1=0$

$\Rightarrow x(\frac{1}{\alpha }+\frac{\lambda }{\beta })+y(\frac{1}{\beta }+\frac{\lambda }{\alpha })-\lambda -1=0$

This meets the axes at $A(\frac{\lambda +1}{\frac{1}{a}+\frac{1}{\beta }},0)$ and $B(0,\frac{\lambda +1}{\frac{1}{\beta }+\frac{\lambda }{\alpha }})$

Let $(h,k)$ be the mid-point of $AB$,

$\Rightarrow h=\frac{1}{2}\cdot \frac{\lambda +1}{\frac{1}{\alpha }+\frac{\lambda }{\beta }},k=\frac{1}{2}\cdot \cdot \frac{\lambda +1}{\frac{1}{\beta }+\frac{\lambda }{\alpha }}$

Eliminating $\lambda$ from these two, we get

$\alpha \beta (h+k)=2(\alpha +\beta )hk$

The locus of $(h,k)$ is $\alpha \beta (x+y)=2(\alpha +\beta )xy$