First we will rearrange the expression from implicit to explicit form. The given equation is

$$(x^2 + 2ay -a^2)^2 = y^2(a^2 - x^2)$$

Squaring the LHS gives

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

Collecting like terms gives the quadratic equation

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

The discriminant is

$$\Delta = {16a^2(x^2 - a^2)^2-4(3a^2+x^2)(x^2-a^2)^2}= (x^2 - a^2)^2(4a^2-x^2) = 4(a^2 - x^2)^3.$$

Solving this equation we get:

$$\eqalign{ y &= {-4a(x^2 - a^2)\pm \sqrt{4(a^2 - x^2)^3}\over 2(3a^2+x^2)} \cr &= {(x^2-a^2)(-2a\pm \sqrt{(a^2-x^2)})\over (3a^2+x^2)} }$$

For each value of $a$ there are two branches of the graph given by taking the +ve and -ve signs in this equation. Values of $y$ are only defined for the interval $-a \leq x \leq a$. The domain of $f(x)$ is $-a \leq x \leq a.$ The zeros of the function are given by $x = a$ and $x = -a.$

Additionally, the function is symmetric with respect to the y-axis, because $x$ always appears in even powers. Graphs for parameters $a = n$ and $a = -n$ and are symmetric to each other with respect to the x-axis, that is $y(n) = y(-n)$. Proof:

$$ {(x^2-n^2)(-2n\pm \sqrt{(n^2-x^2)})\over (3n^2+x^2)} = -{(x^2-(-n)^2)(-2(-n)\pm \sqrt{(-(n)^2-x^2)})\over (3(-n)^2+x^2)}$$

Here are some graphs of the function: