Question

Eliminate $x,y$ between the equations $x^2-y^2=px-qy$, $4xy=qx+py$, $x^2+y^2=1$.

Answer 1

$x^2-y^2=px-qy$ ......... $\left(i\right)$

$4xy=qx+py$......... $\left(ii\right)$

$x^2+y^2=1$......... $\left(iii\right)$

Let's multiply the equation $\left(i\right)$ by $x$ and equation $\left(ii\right)$ by $y,$ we get
$(x^3-y^{2}x)=p{x}^2-qyx$
$4x{y}^2=qxy+p{y}^2$
Add both of them, we get
$x^3+3x{y}^2=p(x^2+y^2)$
But we know, $x^2+y^2=1$
$\therefore x^3+3x{y}^2=p$
Similarly, we get
$y^3+3x^{2}y=q$
Add both of them, we get
$x^3+3x{y}^2+3y{x}^2+y^3=p+q$
Thus , $p+q=(x+y{)}^3,p-q=(x-y{)}^3$
$(p+q{)}^{2/3}+(p-q{)}^{2/3}=(x+y{)}^2+(x-y{)}^2$
$(p+q{)}^{2/3}+(p-q{)}^{2/3}=2(x^2+y^2)$
$(p+q{)}^{2/3}+(p-q{)}^{2/3}=2$