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$$ xy^2 = 1$$

From the equation, it can be said that it is a rectangular hyperbola . Now if I want to calculate the vertex of it , I do follows :

$$\rightarrow (\frac{x}{2}+\frac{y^2}{2})^2 - (\frac{x}{2} - \frac {y^2}{2})^2=1$$ $$\rightarrow X^2-Y^2=1$$

where, $X=\frac{x}{2}+\frac{y^2}{2}$ and $Y=\frac{x}{2}-\frac{y^2}{2}$ .

Now, for vertex : $$X=\pm a,Y=0 (a=1)$$ which implies : $$ \frac{x}{2}+\frac{y^2}{2}=\pm 1 ....(i) $$ $$ \frac{x}{2}-\frac{y^2}{2}=0 ....(ii) $$

Working with $(i)$ and $(ii)$ , I get : $x=\pm 2, y=0$ .

This, vertex is $(\pm 2,0)$ . Am I correct ?

Entrepreneur
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    $xy^2=1$ is not a rectangular hyperbola, or any conic section. The equation of a conic has degree $2$, but $xy^2=1$ has degree $3$. Moreover, $(\pm 2, 0)$ cannot be a vertex, since it does not satisfy the equation: $(\pm 2)(0)^2 = 0 \neq 1$. – Blue Sep 01 '18 at 09:36
  • But the graph seems to be a hyperbola... Or is it only a curve like other 3-degree polynomials? @Blue – Entrepreneur Sep 01 '18 at 09:43
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    Lots of graphs seem to be hyperbolas. :) This one, however, isn't terribly convincing: Its "branches" live in the first and fourth quadrants; a hyperbola's branches would be in opposite quadrants (say, first and third, or second and fourth). What you have is a cubic curve. These have been classified, but the classification is far more elaborate than ellipse/parabola/hyperbola. – Blue Sep 01 '18 at 09:55

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