# How do you identity if the equation x^2-y^2+8x=16 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

Jul 4, 2017

The given equation fits the general Cartesian form of a conic section:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

where $A = 1 , B = 0 , C = - 1 , D = 8 , E = 0 , \mathmr{and} F = - 16$

In the section entitled, Discriminant, the reference tells us to compute ${B}^{2} - 4 \left(A\right) \left(C\right)$:

${0}^{2} - 4 \left(1\right) \left(- 1\right) = 4$

The result is $> 0$, therefore, is it a hyperbola. Furthermore, $A + C = 1 - 1 = 0$, therefore, it is a rectangular hyperbola.

To graph the hyperbola, I recommend that you write the equation in a form that gives you, the center, the vertices and the equations of the asymptotes:

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

center: $\left(h , k\right)$
vertices: $\left(h - a , k\right)$ and $\left(h + a , k\right)$
asymptotes: $y = - \frac{b}{a} \left(x - h\right) + k$ and $y = \frac{b}{a} \left(x - h\right) + k$

Let's work on that form.

Add ${h}^{2}$ to both sides of the equation:

${x}^{2} + 8 x + {h}^{2} - {y}^{2} = 16 + {h}^{2}$

The expansion of the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ tells us that the following equation will allow us to find the value of h:

$- 2 h x = 8 x$

$h = - 4$

The absence of a y term tells us that $k = 0$

Substitute the left side of the pattern into the equation:

${\left(x - h\right)}^{2} - {\left(y - k\right)}^{2} = 16 + {h}^{2}$

Substitute the values for h and k into the equation:

${\left(x - \left(- 4\right)\right)}^{2} - {\left(y - 0\right)}^{2} = 16 + {\left(- 4\right)}^{2}$

Simplify

${\left(x - \left(- 4\right)\right)}^{2} - {\left(y - 0\right)}^{2} = 32$

Divide both sides by 32:

${\left(x - \left(- 4\right)\right)}^{2} / 32 - {\left(y - 0\right)}^{2} / 32 = 1$

Write the denominators as squares:

${\left(x - \left(- 4\right)\right)}^{2} / {\left(4 \sqrt{2}\right)}^{2} - {\left(y - 0\right)}^{2} / {\left(4 \sqrt{2}\right)}^{2} = 1$

The center is $\left(- 4 , 0\right)$

The vertices $\left(- 4 - 4 \sqrt{2} , 0\right)$ and $\left(- 4 + 4 \sqrt{2} , 0\right)$

The equation of the asymptotes are:

$y = - \frac{b}{a} \left(x - h\right) + k$ and $y = \frac{b}{a} \left(x - h\right) + k$

$y = - \left(x - \left(- 4\right)\right)$ and $y = \left(x - \left(- 4\right)\right)$

$y = - x - 4$ and $y = x + 4$

Here is a graph with the equation, the center, the vertices and the asymptotes: 