Question

Solve the following equations:
$2a(\frac{x}{y}-\frac{y}{x})+4a^2=4x^2+\frac{xy}{2a}-\frac{y^2}{a^2}=1$.

Answer 1

$2a(\frac{x}{y}-\frac{y}{x})+4a^2=1⟶\left(I\right)$
$4x^2+\frac{xy}{2a}-\frac{y^2}{a^2}=1⟶\left(II\right)$
$⟹$ Equation $\left(I\right):-$
Let $\frac{y}{x}=p$
$2a(\frac{1}{p}-p)+4a^2=1$
$2a-2a{p}^2+4a^{2}p=p$
$⟹2a{p}^2+p(-4a^2+1)-2a=0$
$⟹p=\frac{-1+4a^2\pm \sqrt{16a^4-8a^2+1+16a^2}}{4a}$
$=\frac{-1+4a^2\pm (-1+4a^2)}{4a}$
$=\frac{8a^2}{4a},\frac{-2}{4a}$
$\frac{y}{x}=2a,-\frac{1}{2a}$
Putting in equation $\left(II\right):-$ $y=-\frac{x}{2a}$
$⟹4x^2-\frac{x\times x}{{\left(2a\right)}^2}-\frac{x^2}{4a^4}=1$
$⟹x^2(4-\frac{1}{4a^2}-\frac{1}{4a^4})=1$
$⟹x^2(16a^4-a^2-1)=4a^4$
$y=2ax$
$4x^2+\frac{x}{2a}\left(2ax\right)-\frac{4a^2{x}^2}{a^2}=1$
$4x^2+x^2-4x^2-1$
$x^2=1$
$\therefore x^2=\frac{4a^4}{{16a}^4-a^2-1}$
$x=\pm \frac{2a^2}{\sqrt{{16a}^4-a^2-1}}$
$y=\frac{x}{2a}$
$y=-\frac{a}{\sqrt{{16a}^4-a^2-1}}$
Solutions are:-
$x=1,y=2a$
$x=-1,y=-2a$
$x=\frac{-2a^2}{\sqrt{{16a}^4-a^2-1}},y=\frac{-a}{\sqrt{{16a}^4-a^2-1}}$
$x=\frac{2a^2}{\sqrt{{16a}^4-a^2-1}},y=\frac{-a}{\sqrt{{16a}^4-a^2-1}}$