Determine if 36x2+4y2216x16y452=0\displaystyle -{36}{x}^{{2}}+{4}{y}^{{2}}-{216}{x}-{16}{y}-{452}={0} is a hyperbola and if it is, which type it is.  The transverse axis is

Horizontal:   (xh)2a2(yk)2b2=1\displaystyle \frac{{\left({x}-{h}\right)}^{{2}}}{{a}^{{2}}}-\frac{{\left({y}-{k}\right)}^{{2}}}{{b}^{{2}}}={1}   or

Vertical:   (yk)2a2(xh)2b2=1\displaystyle \frac{{\left({y}-{k}\right)}^{{2}}}{{a}^{{2}}}-\frac{{\left({x}-{h}\right)}^{{2}}}{{b}^{{2}}}={1}   

If it is a hyperbola, write the equation of the hyperbola in its standard form.  If it is not a hyperbola, leave the rest blank.

h =
k =
a =
b =