what is the directrix and focus of the equation 1/16(y+4)^2=x-3
Perform a substitution
x' = x-3
y' = y+4
and you are left with the equatiom
y'^2 = 16x'
This is the equation of a parabola with axis along the line y = -4 and vertex at (x',y') = (0,0)
which is (x,y) = (3,-4)
The factor 16 tells you where the focus and directrix are.
16 = 4p,
where p is the distance from the vertex to the focus and directrix.
That puts the focus at (x,y) = (7,-4)
The directrix is the vertical line x = -1
To determine the directrix and focus of the equation, let's start by rearranging it in the standard form for a parabola.
1/16(y + 4)^2 = x - 3
Multiply both sides by 16 to eliminate the fraction:
(y + 4)^2 = 16(x - 3)
This equation represents a parabola with its vertex at the point (h, k), where h = 3 and k = -4. It opens to the right since the x-term is positive.
The general equation for a parabola with vertex (h, k) and the axis of symmetry parallel to the x-axis is:
(y - k)^2 = 4a(x - h)
Comparing this general equation to the rearranged equation, we see that a = 4 and h = 3.
The distance from the vertex to the focus (F) is given by the formula:
a = 1/(4F)
Substituting the value of a = 4, we can solve for F:
4 = 1/(4F)
4F = 1/4
F = 1/16
So, the focus (F) of the parabola is located at (1/16, -4).
Similarly, the distance from the vertex to the directrix (D) is also given by the formula:
a = -1/(4D)
Since a = 4, we can find D:
4 = -1/(4D)
4D = -1/4
D = -1/16
Thus, the directrix (D) of the parabola is a vertical line located at x = -1/16.
In summary, the directrix of the given parabola is the vertical line x = -1/16, and the focus is located at the point (1/16, -4).