Question

Find standard form of 3y^2+6y+108x-969=0
The form should be (y-k)^2=4p(x-h)

Answer 1

\[3(y^2+2y+1)=-108x+969+3\]
\[3(y+1)^2=-108x+972\]
\[(y+1)^2=-36x+324\]
\[(y+1)^2=-36(x-9)\]