Solving Systems of equations for $(x,y)\in\mathbb {R}^2$

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So I'm working on solving a couple of system of equations: $$ \text{Let} \ a,b \ \text {be a positive real number with} \ a\neq b \ \text{Solve the system:}$$ $$a^2x^2-2abxy+b^2y^2-2a^2bx-2ab^2y+a^2b^2=0$$$$\text {and}$$$$abx^2+(a^2-b^2)xy-aby^2+ab^2x-a^2by=0$$ I've worked these down to try and pull out some binomials that will cancel out and hopefully shorten the equations down and make them a little nicer to work with.

I'll try to organize this as well as I can.

1) The first thing I did though was graph them and saw that it was a parabola and a hyperbola.

2) Then I also solved for $x$ and $y$ intercepts:

$\qquad\qquad$ 2a)The parabola has $y$-intercepts at $a$ and $x$-intercepts at $b$

$\qquad\qquad$ 2b)The Hyperbola has $y$-intercepts at $-a$ and $x$-intercepts at $-b$

3) I then rearranged the equations to try and simplify things and got: $$(ax-by)^2-2ab(ax+by)+a^2b^2=0$$$$(ax-by)(bx+ay)-ab(bx-ay)=0$$

I am trying to be able to factor things out and reorganize the equations so that I can solve for a, and b, but I'm kind of lost as for what to do now.

Any Ideas would be appreciated!! Thanks!

First Edit:

So I solved the top equation for $x$ and found that $x$ could be $$\frac {(by+ab) \pm 2b\sqrt {ay}}a$$

Then I plugged that back into the bottom equation for $x$ and found that it equaled $$\frac{b^3y^2+2ab^3y+4b^3y\sqrt{ay}+a^2b^3+4ab^3\sqrt{ay}+4b^3ay+a^2by^2+a^3by+2a^2by\sqrt{ay}-b^3y-ab^3-2b^3\sqrt{ay}-a^2by^2-ab^3y-a^2b^3-2ab^3\sqrt {ay}-a^4by}a$$

I'll add more as it comes....

Second Edit

Now I have grouped all of the $\sqrt{ay}$ terms together and moved everything else to the other side so I'll be able to solve for $y$. It's a mess and I'm going to end up with a $y^4$ term just like @Gerry Myerson said. So now I'm going to work on clearing up that mess and see how it simplifies down and see if i can solve the quartic for $y$

Am I on the right track?