# How do you write the equation9x^2 - y^2 - 18x + 6y - 36 = 0 in standard form and what type of conic is it?

Jun 28, 2017

It is a hyperbola.

#### Explanation:

$9 {x}^{2} - {y}^{2} - 18 x + 6 y - 36 = 0$ can be written as

$\left(9 {x}^{2} - 18 x\right) - \left({y}^{2} - 6 y\right) - 36 = 0$

or $9 \left({x}^{2} - 2 x + 1\right) - \left({y}^{2} - 6 y + 9\right) - 36 - 9 + 9 = 0$

or $9 {\left(x - 1\right)}^{2} - {\left(y - 3\right)}^{2} = 36$

or ${\left(x - 1\right)}^{2} / 4 - {\left(y - 3\right)}^{2} / 36 = 1$

As it is of the type ${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1$

it is a hyperbola with center at $\left(1 , 3\right)$

It is a hyperbola.

The graph appears as follows:

graph{9x^2-y^2-18x+6y-36=0 [-20, 20, -6.88, 13.12]}