Tangent line of a lemniscate at (0,0) I need to find the tangent line of the function $y=g(x)$ implicitly defined by
$(x^2+y^2)^2-2a^2(x^2-y^2)=0$
at $(0,0)$, but I don't know how.
I can't use implicit differentiation and evaluate at $(0,0)$, because when $y=0$ I can't use the Implicit Function Theorem to calculate the derivative and, therefore, the slope of the tangent line.
I'd appreciate your help.
Thanks.

$$(x^2+y^2)^2-2a^2(x^2-y^2)=0$$
Solving for $y$ we do substitution $t=y^2$
$$x^4+x^2t+t^2-2a^2x^2+2a^2t=0$$
$$t^2+t(2x^2+2a^2)+x^4-2a^2x^2=0$$
$$t=\pm a\sqrt{4x^2+a^2}-x^2-a^2$$
As $t=-a\sqrt{4x^2+a^2}-x^2-a^2$ is not positive we get solutions
$$y=\pm \sqrt{a\sqrt{4x^2+a^2}-x^2-a^2}$$
Let $f(x)=\sqrt{a\sqrt{4x^2+a^2}-x^2-a^2}$. We have $y=\pm f(x)$, $y'=\pm f'(x)$. The derivative of $f$ is the following
$$f'(x)=\frac{1}{2\sqrt{a\sqrt{4x^2+a^2}-x^2-a^2}}\cdot
\left(
a\frac{1}{2\sqrt{4x^2+a^2}}\cdot(8x)-2x
\right)$$
$$=\frac{x\left(a\frac{2}{\sqrt{4x^2+a^2}}-1\right)}{\sqrt{a\sqrt{4x^2+a^2}-x^2-a^2}}$$
Function $f$ is continuous and differentiable at any small neighborhood of $0$ excluding $0$ itself. The limits of $f'(x)$ as $x\to0^\pm$ exist and they are given by
$$\lim\limits_{x\to0^+}f'(x)=1$$
$$\lim\limits_{x\to0^-}f'(x)=-1$$
Thus at $(0,0)$ there are two tangent lines with equations
$$t_1(x)=x$$
$$t_2(x)=-x$$
