What is the standard form of the equation of the parabola with a focus at (16,-3) and a directrix of #y=3 #?

Question

Kennedy Adams · Accepted Answer

#y=-1/12*(x^2-32*x+256)# The vertex is the mid-point between focus and directrix. So the vertex is at #(16,0)#. The equation ofv the parabola is #y=a(x-16)^2+0# The distance from Vertex to focus and directrix is #3=1/(4|a|) ; a = -1/12 # (- sign as the parabola opens down here) #:. y=-1/12*(x^2-32*x+256)# graph{-1/12(x-16)^2 [-40, 40, -20, 20]}[Ans]

Kennedy Adams · Answer

undefined The standard form of the equation of a parabola with a vertical axis of symmetry is given by:

$$ (x - h)^2 = 4p(y - k) $$

where:
- (h, k) is the vertex of the parabola.
- p is the distance from the vertex to the focus (or from the vertex to the directrix if the parabola opens downward).
- If the parabola opens upward, p is positive; if it opens downward, p is negative.

Given that the focus is at (16, -3) and the directrix is $ y = 3 $, we can see that the parabola opens downward since the focus is below the directrix.

First, we find the vertex, which is the midpoint between the focus and the directrix. The y-coordinate of the vertex is the average of the y-coordinate of the focus and the y-coordinate of the directrix.

$$ y_{\text{vertex}} = \frac{(-3 + 3)}{2} = 0 $$

Since the parabola opens downward, the vertex is the highest point, so we choose the focus as the vertex.

Now, the distance from the vertex to the focus (p) is the absolute value of the difference between the y-coordinate of the vertex and the y-coordinate of the focus.

$$ p = |-3 - 3| = 6 $$

So, the vertex is (16, -3) and $ p = 6 $.

Plugging these values into the standard form equation:

$$ (x - 16)^2 = -24(y + 3) $$

Thus, the standard form of the equation of the parabola is $ (x - 16)^2 = -24(y + 3) $.