find the vertex,focus and equation of the direction of y=-1/16(x-6)^2+3
Y=1/2p(x-h)^2+k
p=8 h=6 k=3
(6,3)
so vertex is (6,3), focus is (6,-1) and directrix = y=7
Is this correct?
1/2 p = -1/16 ... so p = ?
vertex is correct
so would y=-1 and F(6,7)
Yes, your answer is correct.
To find the vertex and the equation of the parabola, we need to write the given equation in the standard form:
y = 1/2p(x - h)^2 + k
Comparing this with the given equation y = -1/16(x - 6)^2 + 3, we can identify the following values:
p = -1/16 (This value is negative because the parabola opens downwards)
h = 6
k = 3
Substituting these values into the standard form equation, we get:
y = 1/2(-1/16)(x - 6)^2 + 3
Simplifying further, we have:
y = -1/32(x - 6)^2 + 3
Therefore, the vertex of the parabola is at point (6, 3), and the equation of the parabola is y = -1/32(x - 6)^2 + 3.
To find the focus and the directrix of the parabola, we need to determine the value of p.
For a parabola in the form y = 1/4p(x - h)^2 + k, the focus is located at the point (h, k + 1/4p), and the directrix is given by the equation y = k - 1/4p.
In our case, p is equal to -1/16.
So, the focus is located at point (6, 3 + 1/4(-1/16)), which simplifies to (6, 3 - 1/64) or approximately (6, -1).
The directrix is given by the equation y = 3 - 1/4(-1/16), which simplifies to y = 3 + 1/64 or approximately y = 7.
Therefore, the focus of the parabola is at point (6, -1), and the equation of the directrix is y = 7.