Hyperbola: Asymptotes

Step 1: Group the x- and y-terms on the left-hand side of the equation.

3x² - 2y² + 18x + 15 = 0

$(3x^2+18x)+(-2y^2)+15=0$

Step 2: Move the constant term to the right-hand side.

$(3x^2+18x)+(-2y^2)=-15$

Step 3: Complete the square for the x- and y-groups.

$(3x^2+18x)+(-2y^2)=-15$

Complete the square for the x-group

(3x² + 18x)

Factor out a 3 so the x²-coefficient is 1

3(x² + 6x)

Take the coefficient of the x-term, divide by 2 and square the result.

$\frac{6}{2}=3;3^2=9$

Add the result to the x-group.

3(x² + 6x + 9)

Complete the square for the y-group

-2y²

Because there is no y-term no work is necessary.

Final result.

$3\left(x^2+6x+9\right)-2y^2=-15+?$

Step 4: Add the values added to the left-hand side to the right-hand side.

$3\left(x^2+6x+9\right)-2y^2=-15+3\left(9\right)$

$3\left(x^2+6x+9\right)-2y^2=12$

Step 5: Write the x-group and y-group as perfect squares.

$3{\left(x+3\right)}^2-2y^2=12$

Step 6: Divide both sides by the value on the right-hand side, simplify and write the denominators as perfect squares.

$\frac{3{\left(x+3\right)}^2}{12}-\frac{2y^2}{12}=1$

Divide by 12:

$\frac{{\left(x+3\right)}^2}{4}-\frac{y^2}{6}=1$

Write denominators as perfect squares:

$\frac{{\left(x+3\right)}^2}{2^2}-\frac{y^2}{{\sqrt{6}}^2}=1$

Step 7: Identify h, k, a and b and the orientation of the transverse axis from the standard equation in Step 6.

Center: (h, k) = (-3, 0)

Note that x + 3 = x - (-3)

a = 2

$b=\sqrt{6}$

The transverse axis is horizontal since x is in the numerator above a².

Step 8: Write the equations of the asymptotes

Equation for a horizontal transverse axis:

$y=0\pm \frac{b}{a}\left(x-k\right)$

Substitute:

$y=0\pm \frac{\sqrt{6}}{2}\left[x-\left(-3\right)\right]$

Simplify:

$y=\pm \frac{\sqrt{6}}{2}\left(x+3\right)$

Step 1: Find the center coordinates.

Center: The center is the midpoint of the two vertices.

$\left(h, k\right)=\left(\frac{4+4}{2}, \frac{3+9}{2}\right)=\left(\frac{8}{2},\frac{12}{2}\right)=\left(4, 6\right)$

Step 2: Determine the orientation of the transverse axis and the distance between the center and the vertices (a).

Orientation of the transverse axis: Since both vertices fall on the vertical line x = 4, the transverse axis is vertical.

Length of a: To find the length between the center and the vertices take either vertices and find the change in their y-coordinates.

Vertex (4, 3): $a=\left|6-3\right|=\left|3\right|=3$

Vertex (4, 9): $a=\left|6-9\right|=\left|-3\right|=3$

a = 3

Step 3: Determine the value of b.

The given asymptote equation, $y=4\pm 2x-12$ has a slope of 2. Because the transverse axis is vertical, $2=a/b$.

$b=\frac{a}{2}=\frac{3}{2}\to b=1.5$

Step 4: Write the standard form of the hyperbola.

Equation for a vertical transverse axis:

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

Substitute:

$\frac{{\left(y-6\right)}^2}{3^2}-\frac{{\left(x-4\right)}^2}{{\left(1.5\right)}^2}=1$

Final equation:

$\frac{{\left(y-6\right)}^2}{3^2}-\frac{{\left(x-4\right)}^2}{{\left(1.5\right)}^2}=1$