Find the fundamental rectangle of a hyperbola

Find the vertices of the fundamental rectangle of hyperbola y=sqrt(36x^2+1)

thanks :)

y=√(36x^2+1)
y² = 36x²+1
y²- 36x² = 1

y² /1 - x² /(1/36) =1
a² =1
b² =1/36


The four corners of the fundamental rectangle of the hyperbola which opens right and left are
(a,b), (a,-b), (-a,b),(-a,-b)