Hi again. I am supposed to graph y+3=-1/12(x-1)^2. How do I find the equation of the directrix, and the coordinates of the vertex and the focus?
To find the equation of the directrix, as well as the coordinates of the vertex and the focus, we can start by putting the given equation in the standard form for a parabola with its vertex at the origin, which is given by:
y = 4p(x - h)^2
where (h, k) is the vertex coordinates and p is the distance from the vertex to the focus and the directrix.
In the given equation, y + 3 = -1/12(x - 1)^2, we can rewrite it as:
y = -1/12(x - 1)^2 - 3
Now, let's simplify this equation further:
y = -1/12(x^2 - 2x + 1) - 3
= -1/12x^2 + 1/6x - 1/12 - 3
= -1/12x^2 + 1/6x - 1/12 - 36/12
= -1/12x^2 + 1/6x - 37/12
Comparing this to the standard form equation, y = 4p(x - h)^2, we can see that the parabola's vertex is (h, k) = (0, -37/12).
To determine the value of p and find the focus and directrix, we need to use the formula:
p = 1/4a
where a is the coefficient of x^2. In this case, a = -1/12, so:
p = 1 / (4 * (-1/12))
To simplify this, we multiply numerator and denominator by 12:
p = 1 / (-1/3)
p = -3
Therefore, the value of p is -3.
Now that we have the value of p, we can find the focus and directrix. For a parabola with a vertical axis (like this one), the focus is located at the point (0, p) and the directrix is a horizontal line given by the equation y = -p.
Plugging in the value of p = -3, we find that the focus is at (0, -3), and the directrix is the horizontal line y = 3.
So, in summary:
- The vertex of the parabola is (0, -37/12).
- The focus is located at (0, -3).
- The equation of the directrix is y = 3.