x^2 - 3y^2 - 8x + 12y + 16 = 0
What is the center and the domain of the equation?
4*y1
What does that mean? I don't get it.
To determine the center and domain of the equation x^2 - 3y^2 - 8x + 12y + 16 = 0, we need to first rewrite the equation in a specific form. The given equation is in the form of a general conic equation, specifically an equation of a hyperbola.
1. Rearrange the equation: x^2 - 8x - 3y^2 + 12y = -16
2. Complete the square for both the x and y terms:
To complete the square for the x terms, take half of the coefficient of x (-8), square it, and add it to both sides of the equation:
x^2 - 8x + 16 - 3y^2 + 12y = -16 + 16
(x - 4)^2 - 3y^2 + 12y = 0
To complete the square for the y terms, take half of the coefficient of y (12), square it, and add it to both sides of the equation:
(x - 4)^2 - 3(y^2 - 4y + 4) = 0
(x - 4)^2 - 3(y - 2)^2 + 12 = 0
3. Bring the equation to standard form:
(x - 4)^2 - 3(y - 2)^2 = -12
Comparing this equation with the standard form of a hyperbola:
(x - h)^2/a^2 - (y - k)^2/b^2 = 1 (for a horizontal hyperbola)
(y - k)^2/a^2 - (x - h)^2/b^2 = 1 (for a vertical hyperbola)
We can deduce the center, domain, and nature of the given hyperbola:
Center: The center of the hyperbola is given by the values (h, k) in the equation. In this case, the center is (4, 2).
Domain: The domain of a hyperbola refers to the range of valid x-values. For a horizontal hyperbola, the domain is all real numbers greater than or less than the x-coordinate of the center. So, the domain of this hyperbola is (-∞, ∞).
Therefore, the center of the given hyperbola is (4, 2), and the domain is (-∞, ∞).