Question

If $x^2+y^2-xy=3$ and $y-x=1$ then find $\frac{xy}{x^2+y^2}$.

Answer 1

We have $y-x=1$ or $y=1+x$

Substitute $y=1+x$ in the equation we get

$x^2+y^2-xy=3$

$\Rightarrow x^2+{(1+x)}^2-x(1+x)=3$

$\Rightarrow x^2+x^2+2x+1-x-x^2-3=0$

$\Rightarrow x^2+x-2=0$

$\Rightarrow (x-1)(x+2)=0$

$\therefore x=1,-2$

When $x=1\Rightarrow y=1+x=1+1=2$

When $x=-2\Rightarrow y=1+x=1-2=-1$

$\therefore \{1,2\}$ and $\{-2,-1\}$ are the solutions of the above equations.

$\therefore {\left[\frac{xy}{x^2+y^2}\right]}_{(1,2)}=\frac{1\times 2}{1+4}=\frac{2}{5}$

$\therefore {\left[\frac{xy}{x^2+y^2}\right]}_{(-2,-1)}=\frac{-2\times -1}{1+4}=\frac{2}{5}$