Question

trace curve of x²y²=a²(x²-y²)​

Answer 1

Answer:

Solving for x2x2 and y2y2 we have

x2=a2y2a2+y2y2=a2x2a2−x2x2=a2y2a2+y2y2=a2x2a2−x2

In the first, we have

limy→∞x2=a2limy→∞x2=a2

and in the second

limx±→±a=±∞limx±→±a=±∞

so the asymptotes are located at x=±ax=±a