# How do you write  6x^2 - 24x - 5y^2 - 10y - 11 = 0 in standard form?

May 21, 2015

The standard form of a hyperbola with a transverse axis (all stuff this is, trust me on this) is
${\left(x - h\right)}^{2} / \left({a}^{2}\right) - {\left(y - k\right)}^{2} / {b}^{2} = 1$

$6 {x}^{2} - 24 x - 5 {y}^{2} - 10 y - 11 = 0$

move he constant to the left side

$6 \left({x}^{2} - 4 x\right) - 5 \left({y}^{2} + 2 y\right) = 11$

complete the squares

$6 \left({x}^{2} - 4 x + 4\right) - 5 \left({y}^{2} + 2 y + 1\right) = 11 + 24 - 5$

$6 {\left(x - 2\right)}^{2} - 5 {\left(y + 1\right)}^{2} = 30$

Divide by 30 so left side equals 1

${\left(x - 2\right)}^{2} / {\left(\sqrt{5}\right)}^{2} - {\left(y + 1\right)}^{2} / {\left(\sqrt{6}\right)}^{2} = 1$