Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$.

Question

Problem: Finding the asymptotes of $(y-a)^2(x^2-a^2)=x^4+a^4$.

My efforts: $(y-a)^2(x^2-a^2)=x^4+a^4\implies (y-a)^2=\dfrac{x^4+a^4}{x^2-a^2}\implies x=\pm a$ are vertical asymptotes. What are the remaining asymptotes?

Answer 1

Rewrite the equation as

$$y=a \pm \sqrt{\frac{x^4+a^4}{x^2-a^2}}$$

The vertical asymptotes $x=\pm a $ are directly identified. Since

$$\lim_{x^2\to\infty} y(x) = a\pm x$$

$y=a\pm x $ are the slant asymptotes.

Answer 2

This doesn't satisfy the usual definition of asymptote but ...

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

$\begin{array}\\ (y-a)^2 &=\dfrac{x^4+a^4}{x^2-a^2}\\ &=\dfrac{x^4-a^2x^2+a^2x^2+a^4}{x^2-a^2}\\ &=\dfrac{x^2(x^2-a^2)+a^2(x^2+a^2)}{x^2-a^2}\\ &=x^2+\dfrac{a^2(x^2-a^2+2a^2)}{x^2-a^2}\\ &=x^2+a^2+\dfrac{2a^4}{x^2-a^2}\\ \text{so}\\ y-a &=\sqrt{x^2+a^2+\dfrac{2a^4}{x^2-a^2}}\\ &=x\sqrt{1+a^2/x^2+\dfrac{2a^4}{x^2(x^2-a^2)}}\\ &\approx x(1+a^2/(2x^2)+\dfrac{2a^4}{2x^2(x^2-a^2)}) \qquad\text{for large }x\\ &= x+a^2/(2x)+O(1/x^3)\\ \end{array} $

so $y = x+a$ is a "asymptote".