Question 847862
put into standard form 
9x^2-16y^2+54x+160y-320=0
9x^2+54x-16y^2+160y-320=0
complete the square:
9(x^2+6x+9)-16(y^2-10y+25)=320+81-400
9(x+3)^2-16(y-5)^2=1
{{{(x+3)^2/(1/9)-(y-5)^2/(1/16)=1}}}
This is an equation of a hyperbola with vertical transverse axis.
Its standard form: {{{(x-h)^2/a^2-(y-k)^2/b^2=1}}}, (h,k)=(x,y) coordinates of center