Convert this in to the standard form of an ellipse

Question

$$ \frac{p^2+q^2a^2}{a^2}-\frac{\left(2px+2qa^2y-a^2\right)}{a^2y^2+x^2}=k^2$$

How to put the above equation into this form? $$\frac{\left(X-H\right)^2}{A^2}+\frac{\left(Y-K\right)^2}{B^2}=1$$

The algebra is killing me :) Any help would be appreciated!

I've tried my hardest and well I'm stuck with this... $$p^2x^2+q^2a^2x^2-k^2a^2x^2+p^2a^2y^2+q^2a^4y^2-k^2a^4y^2-2pa^2x-2qa^4x+a^4=0$$ really don't know what to do from here on.

Answer 1

Let $p^2+q^2a^2-k^2a^2=C$ (I used $C^2$ in the comments for 'dimensional' reasons))

to obtain $$Cx^2-2a^2px+a^2(Cy^2-2a^2qy)+a^4=0$$

You can then multiply through by $C$ and complete the square to obtain $$(Cx-a^2p)^2+a^2(Cy-a^2q)^2=a^4(p^2+a^2q^2-C)=k^2a^6$$

And you should be able to complete it from there.

Answer 2

Hint:

Regroup the terms with equal powers of $x$ and $y$ (actually, you just have term in $x^2,x,y^2,1$ which isn't that much).

The standard form of the ellipse equation is the sum of two squares, the one generating terms in $x^2,x,1$, and the other $y^2,y,1$.

So it seems that you can just handle $x$ and $y$ separately for square completion.

$$a(x+b)^2+c(y+d)^2+e=ax^2+(2ab)x+cy^2+(2cd)y+(b^2+d^2+e).$$

Note that the lack of an $y$ term is not an obstacle to square completion, on the opposite.