Find the area bounded by $x^2y^2+y^4-x^2-5y^2+4=0.$

Question

Find the area bounded by $x^2y^2+y^4-x^2-5y^2+4=0.$

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I reduced the above equation to $y^2=\frac{x^2-4}{x^2+y^2-5}$ but i am not able to solve further.

Answer 1

Solve with respect to $x^2$ and obtain, for $y\ne \pm 1$, then $$ x^2=-\frac{y^4-5y^2+4}{y^2-1}=4-y^2. $$ Hence, circle of area $4\pi$.

Note. The points $(x,\pm 1)$, $x\in\mathbb R$, also belong to the $$ x^2y^2+y^4-x^2-5y^2+4=0. $$

Answer 2

This equation is equivalent to $$ \left(x^2+y^2-4\right)\left(y^2-1\right)=0 $$ That is, this describes the union of a circle of radius $2$ centered at the origin and a horizontal line at $y=1$ and a horizontal line at $y=-1$. So, the enclosed area is inside the circle, which has area $4\pi$.