What type of conic section has the equation 4x^2 - y^2 +4y - 20= 0?

Sep 25, 2014

$4 {x}^{2} - {y}^{2} + 4 y - 20 = 0$

$4 {x}^{2} - {y}^{2} + 4 y = 20$

$4 {\left(x - 0\right)}^{2} - {y}^{2} + 4 y = 20$

We need to complete the square using the $y$ term.

${\left(\frac{4}{2}\right)}^{2} = {\left(2\right)}^{2} = 4$, Add $4$ to both sides

$4 {\left(x - 0\right)}^{2} - {y}^{2} + 4 y + 4 = 20 + 4$

${y}^{2} + 4 y + 4 \implies {\left(y + 2\right)}^{2} \implies$ Perfect square trinomial

Re-write:

$4 {\left(x - 0\right)}^{2} - {\left(y + 2\right)}^{2} = 20 + 4$

$4 {\left(x - 0\right)}^{2} - {\left(y + 2\right)}^{2} = 24$

$\frac{4 {\left(x - 0\right)}^{2}}{24} - \frac{{\left(y + 2\right)}^{2}}{24} = \frac{24}{24}$

$\frac{{\left(x - 0\right)}^{2}}{6} - \frac{{\left(y + 2\right)}^{2}}{24} = 1 \implies$Hyperbola