solution:
quadratic surface,
x^{2}+z^{2}+\frac{y^{2}}{36}=1
=x^{2}+z^{2}=1-\frac{y^{2}}{36}
=\frac{x^{2}}{a^{2}}+\frac{z^{2}}{c^{2}}=1-\frac{y^{2}}{b^{2}}
this is a hyperbolid of one that intersection at
y=0
the quadratic surface becomes
x^{2}+z^{2}=1
this is the equation of the unit circle
so,the surface is three circle
