Question

change y^2-4x^2-16x-2y-19=0 to standard form

Answer 1

\[\begin{align*}
y^2-2y-4x^2-16x-19&=0\\
\left(y^2-2y+1\right)-4\left(x^2+4x\right)-20&=0\\
\left(y-1\right)^2-4\left(x^2+4x+4-4\right)-20&=0\\
\left(y-1\right)^2-4\left((x+2)^2-4\right)-20&=0\\
\left(y-1\right)^2-4(x+2)^2+16-20&=0\\
\left(y-1\right)^2-4(x+2)^2&=4\\
\frac{\left(y-1\right)^2}{4}-(x+2)^2&=1\\
\end{align*}\]