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?

1 Answer
Jul 4, 2017

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

#Ax^2+Bxy+Cy^2+Dx+Ey+F = 0#

where #A = 1, B=0, C = -1, D=8, E = 0, and F = -16#

In the section entitled, Discriminant, the reference tells us to compute #B^2-4(A)(C)#:

#0^2-4(1)(-1) = 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:

#(x^2-h)^2/a^2-(y-k)^2/b^2=1#

center: #(h,k)#
vertices: #(h-a,k)# and #(h+a,k)#
asymptotes: #y = -b/a(x-h)+k# and #y = b/a(x-h)+k#

Let's work on that form.

Add #h^2# to both sides of the equation:

#x^2+8x+h^2-y^2=16+h^2#

The expansion of the pattern #(x-h)^2=x^2-2hx+h^2# tells us that the following equation will allow us to find the value of h:

#-2hx = 8x#

#h = -4#

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

Substitute the left side of the pattern into the equation:

#(x-h)^2-(y-k)^2=16+h^2#

Substitute the values for h and k into the equation:

#(x-(-4))^2-(y-0)^2=16+(-4)^2#

Simplify

#(x-(-4))^2-(y-0)^2=32#

Divide both sides by 32:

#(x-(-4))^2/32-(y-0)^2/32=1#

Write the denominators as squares:

#(x-(-4))^2/(4sqrt2)^2-(y-0)^2/(4sqrt2)^2=1#

The center is #(-4,0)#

The vertices #(-4-4sqrt2,0)# and #(-4+4sqrt2,0)#

The equation of the asymptotes are:

#y = -b/a(x-h) + k# and #y = b/a(x-h)+k#

#y = -(x- (-4))# and #y = (x- (-4))#

#y = -x-4# and #y = x+4#

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