graphing a hyperbola.
-4x^2+9y^2+8x+72y-4=0
- just need help setting it up into an equation
To graph a hyperbola, you first need to rewrite the given equation in a standard form. The standard form of a hyperbola equation is:
((x - h)^2 / a^2) - ((y - k)^2 / b^2) = 1 (for a horizontal hyperbola) or
((y - k)^2 / a^2) - ((x - h)^2 / b^2) = 1 (for a vertical hyperbola)
where (h, k) represents the center of the hyperbola, and a and b are the lengths of the transverse and conjugate axes.
Let's rewrite the given equation to match the standard form:
-4x^2 + 9y^2 + 8x + 72y - 4 = 0
First, let's group the x terms and y terms together:
(-4x^2 + 8x) + (9y^2 + 72y) - 4 = 0
Next, complete the square for both the x and y terms. To complete the square for the x terms, we factor out -4 from the first two terms:
-4(x^2 - 2x) + (9y^2 + 72y) - 4 = 0
Now, we need to find the value to add to complete the square for the x terms. To find this value, take half of the coefficient of the x term (in this case, (-2 / 2 = -1)), square it (1^2 = 1), and add it to both sides of the equation:
-4(x^2 - 2x + 1) + (9y^2 + 72y) - 4 + 4 = 1
Now, complete the square for the y terms. To do this, we take half of the coefficient of the y term (in this case, (72 / 2 = 36)), square it (36^2 = 1296), and add it to both sides of the equation:
-4(x^2 - 2x + 1) + (9y^2 + 72y + 1296) - 4 + 4 = 1 + 1296
Now, let's simplify the equation:
-4(x - 1)^2 + 9(y + 36)^2 = 1301
Finally, divide both sides of the equation by the constant term (1301) to have the equation equal to 1:
-4(x - 1)^2 / 1301 + 9(y + 36)^2 / 1301 = 1
Therefore, the standard form of the hyperbola equation is:
((x - 1)^2 / (1301/4)) - ((y + 36)^2 / (1301/9)) = 1
From this equation, we can identify the center of the hyperbola as (1, -36), and the lengths of the transverse and conjugate axes as the square roots of (1301/4) and (1301/9), respectively.