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

Answer:

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.