# How do I graph the hyperbola with the equation 4x^2−25y^2−50y−125=0?

Mar 2, 2015

We have to write it in the standard form:

$4 {x}^{2} - 25 {y}^{2} - 50 y - 125 = 0 \Rightarrow$

$4 {x}^{2} - 25 \left({y}^{2} + 2 y\right) = 125 \Rightarrow$

$4 {x}^{2} - 25 \left({y}^{2} + 2 y + 1 - 1\right) = 125 \Rightarrow$

$4 {x}^{2} - 25 {\left(y + 1\right)}^{2} + 25 = 125 \Rightarrow$

$4 {x}^{2} - 25 {\left(y + 1\right)}^{2} = 100 \Rightarrow$

$\frac{4}{100} {x}^{2} - \frac{25}{100} {\left(y + 1\right)}^{2} = 1 \Rightarrow$

${x}^{2} / 25 - {\left(y + 1\right)}^{2} / 4 = 1$

This is an hyperbola centered in $C \left(0 , - 1\right)$, with the branches left and right (there is an $1$ at the second member), with the asymptotes:

$m = \pm \frac{4}{25}$, so:

$y - \left(- 1\right) = \pm \frac{4}{25} \left(x - 0\right) \Rightarrow y = \pm \frac{4}{25} x - 1$.

graph{4x^2-25y^2-50y-125=0 [-10, 10, -5, 5]}