How do you identity if the equation 2x^2+12x+18-y^2=3(2-y^2)+4y is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Nov 15, 2016

Set your compass to a radius of 2, put the center at the point (-3, 1), and draw a circle.

Explanation:

Here is a helpful reference Conic Section - General form

Combine all of the terms to be in the general form:

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

2x^2 + 0xy + 2y^2 + 12x - 4y + 12 = 0" [1]"

Because A = C and B = 0, this is an equation of a circle. The general equation for a circle is:

(x - h)^2 + (y - k)^2 = r^2

where (x,y) is any point on the circle, (h, k) is the center, and r is the radius.

Divide both sides of equation [1] by 2:

x^2 + y^2 + 6x - 2y + 6 = 0" [2]"

Add h^2 + k^2 - 6 to both sides of the equation:

x^2 + 6x + h^2 + y^2 - 2y + k^2 = h^2 + k^2 - 6" [3]"

Set the term middle in the right side of the pattern, (x - h)^2 = x^2 -2xh + h^2, equal to the corresponding term in equation [3]:

-2hx = 6x

Solve for h:

h = -3

Substitute the left side on the pattern into equation 3:

(x - h)^2 + y^2 - 2y + k^2 = h^2 + k^2 - 6" [4]"

Substitute -3 for h in equation [4]:

(x - -3)^2 + y^2 - 2y + k^2 = (-3)^2 + k^2 - 6" [5]"

Set the term middle in the right side of the pattern, (y - k)^2 = y^2 -2yk + k^2, equal to the corresponding term in equation [5]:

-2yk = -2y

Solve for k:

k = 1

Substitute the left side of the pattern into equation [5]:

(x - -3)^2 + (y - k)^2 = (-3)^2 + k^2 - 6" [6]"

Substitute 1 for k into equation [7]:

(x - -3)^2 + (y - 1)^2 = (-3)^2 + 1^2 - 6" [7]"

Simplify the constant terms:

(x - -3)^2 + (y - 1)^2 = 4" [8]"

Write the constant term as a square:

(x - -3)^2 + (y - 1)^2 = 2^2" [9]"

Equation [9] is the standard form of a circle with a center at (-3, 1) and a radius of 2.