Finding the Inverse of a Nested Square-Root Function

Question

I'm a CS grad student working on a project where one of the issues at the moment comes from trying to find the inverse to the following function:

$f(x)=\sqrt{-a-x^2+2\sqrt{a x^2+x^4}}$

I already know that this function only has real values (and therefore a useful inverse) when $x^2\geq\frac{a}{3}$ but I haven't been able to make any real progress on the inverse itself. The graph of this looks an awful lot like a hyperbola, but it doesn't seem to follow any standard form (eg. $1=3\frac{x^2-y^2}{a}$).

Any help on this would be greatly appreciated.

Answer 1

after squaring you will get $$y^2+a+x^2=2\sqrt{ax^2+x^4}$$ squaring again we get $$(y^2+a)^2+x^4+2x^2(y^2+a)=4ax^2+4x^4$$ or $$-3x^4+x^2(2y^2-2a)+(y^2+a)^2=0$$ now set $$t=x^2$$

Answer 2

Depending on how flexible your definition of "standard form" is, the following might be suitable: Squaring both sides of the equation $y = f(x)$, moving $-(a+x^2)$ to the other side of the equation, and squaring again shows that every point $(x,y) = (x,f(x))$ lies on the variety

$$ (x^2 + y^2 - a)^2 = 4x^2(a + x^2). $$

We can rewrite this equation as

$$ -3x^4 + 2(y^2 - 3a)x^2 + (y^2-a)^2 = 0, $$

and so, solving for $x^2$ with the quadratic formula and simplifying,

$$ |x| = \pm\sqrt{\frac{1}{6}\left(2(y^2-3a) \pm 4\sqrt{y^4-3ay^2+3a^2}\right)}. $$

You can then, either analytically or by testing points, determine which choices of signs give you the $(x,y)$ pairs corresponding to $y = f(x)$.