Question

Eliminate $x,y$ from the equations $a{x}^2+b{y}^2=ax+by=\frac{xy}{x+y}=c$.

Answer 1

We have

$c(a{x}^2+b{y}^2)=c^2=-{(ax+by)}^2$

$\therefore a(a-c)x^2+2abxy+b(b-c)y^2=0⟶1$

$(ax+by)(x+y)=xy;$

$\therefore a{x}^2+(a+b-1)xy+b{y}^2=0⟶2$

From $1$ & $2$ by cross multiplication

$=\frac{x^2}{b\{2ab-(b-c\left)\right(a+b-1\left)\right\}}=\frac{xy}{-ab(a-b)}=\frac{y^2}{a\left\{\right(a-c\left)\right(a+b-1)-2ab\}}$

$\therefore \{2ab-(b-c)(a+b-1)\}\{(a-c)(a+b-1)-2ab\}=ab{(a-b)}^2;$

$\therefore 2ab(a+b-1)(a+b-2c)-4a^2{b}^2-(a-c)(b-c)(a+b-1)=ab{(a-b)}^2$

$\therefore 2ab(a+b)(a+b-1)-4abc(a+b-1)-4a^2{b}^2-ab{(a+b-1)}^2+c(a+b){(a+b-1)}^2-c^2{(a+b-1)}^2=ab{(a-b)}^2$

Where by arranging according to powers of 'c', we have

$ab(a+b-1)(a+b-1)+c(a+b-1)\{{(a-b)}^2-(a+b)\}-c^2{(a+b-1)}^2=ab{(a-b)}^2$

i.e, $c^2{(a+b-1)}^2-c(a+b-1)\{{(a-b)}^2-(a+b)\}+ab=0$