Question

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?

Answer 1

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.