What are the vertex, focus, and directrix of the parabola with the given equation?
12y = x2 –6x + 45
(1 point)
The given equation is in the form of y = ax^2 + bx + c, where a = 1, b = -6, and c = 45.
To find the vertex, we use the formula x = -b/(2a).
Substituting the values, we have x = -(-6)/(2*1) = 6/2 = 3.
Therefore, the x-coordinate of the vertex is 3.
To find the y-coordinate of the vertex, we substitute x = 3 into the given equation:
12y = (3)^2 – 6(3) + 45
12y = 9 – 18 + 45
12y = 36
y = 36/12
y = 3.
Therefore, the y-coordinate of the vertex is 3.
Hence, the vertex is (3, 3).
To find the focus, we use the formula (h + p, k), where (h, k) is the vertex and p = 1/(4a).
Substituting the values, we have p = 1/(4*1) = 1/4.
Therefore, the focus is at (3 + 1/4, 3) = (13/4, 3).
To find the directrix, we use the equation x = h – p.
Substituting the values, we have x = 3 – 1/4 = 12/4 - 1/4 = 11/4.
Therefore, the directrix is x = 11/4.