# How do you write x^2 - 8y^2 - 14x -15 = 0 in standard form?

May 21, 2015

This is a hyperbola with a horizontal transverse axis so its standard form is

${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

Given ${x}^{2} - 8 {y}^{2} - 14 x - 15 = 0$

${x}^{2} - 14 x + 49 - 8 {y}^{2} = 15 + 49$

${\left(x - 7\right)}^{2} - 8 {y}^{2} = 64$

${\left(x - 7\right)}^{2} / {8}^{2} - {y}^{2} / {\left(2 \sqrt{2}\right)}^{2} = 1$