Find all asymptotes of given curve

Question

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

origin i found will be a node

i'm trying to find asymptotes by equating highest degree term to $0$

Please help.

Answer 1

write $y(x)=mx+b$ and then rearrange equation in the form $\sum_0^n a_nx^n=0$, where n is the highest degree term you get once you rearrange equation. Now you set $a_n$ and $a_{n-1}$ to $0$ and for each pair {m,b} found you get an asymptote. in your case if i didn't make any mistakes you should find only one asymptote at $x=-a$

Answer 2

Consider $y^2(a+x) = x^2(b-x) $, with $a, b \ne 0$.

Since $(0, 0)$ is a point on this curve, suppose $y'(0) = c$.

Note that this is more general.

Then, for small $x$ and $y$, $y \approx cx$.

Then $(cx)^2(a+x) \approx x^2(b-x) $ or $c^2(a+x) \approx b-x $ or, since $x$ is small, $c^2a \approx b$ or $c^2 \approx b/a$.

Therefore $c = \pm \sqrt{b/a}$ so there are two asymptotes at the origin with slopes $\pm \sqrt{b/a} $.

In your case, since $a=b$, the slopes are $\pm 1$.