# How do you identity if the equation (x-1)^2-9(y-4)^2=36 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Oct 25, 2016

It is a hyperbola.
Its center is $\left(1 , 4\right)$
Its vertices are :(7, 4) ; (-5, 4)
Its foci $\left(1 \pm 2 \sqrt{10} , 4\right)$
Asymptots; $y = \pm \frac{2}{3} x + \frac{10}{3}$

#### Explanation:

Given -

$\left(x - 1\right) 2 - 9 {\left(y - 4\right)}^{2} = 36$

It is a hyperbola.

We shall write it in the standard form -

$\frac{x - 1}{36} - \frac{9 {\left(y - 4\right)}^{2}}{36} = \frac{36}{36}$
$\frac{x - 1}{36} - \frac{y - 4}{4} = 1$

It is in the form

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

Its center is $\left(1 , 4\right)$

Since the $\frac{x - 1}{36}$term is positive it opens right and left

Its vertices are :

$\left(x + a , y\right)$
${a}^{2} = 36$
$a = 6$
$\left(1 + 6 , 4\right)$
$\left(7 , 4\right)$
$\left(x - a , y\right)$
$\left(1 - 6 , 4\right)$
$\left(- 5 , 4\right)$

Its foci are -

$f = \sqrt{{a}^{2} + {b}^{2}}$
$f = \sqrt{36 + 4} = \sqrt{40}$
$f = \sqrt{40} = 2 \sqrt{10}$

$\left(1 \pm 2 \sqrt{10} , 4\right)$

Asymptots

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$m =$Rise/run$\frac{\Delta y}{\Delta x} = \frac{4}{6} = \frac{2}{3}$

$y - 4 = \frac{2}{3} \left(x - 1\right)$

$y = \frac{2}{3} x - \frac{2}{3} + 4$

$y = \pm \frac{2}{3} x + \frac{10}{3}$