Question

In order to successfully perform a trick, a flying trapeze artist must swing along a parabolic path that is equidistant from the floor and the pivot point where the trapeze rope is attached. The rope is attached to the ceiling 8 feet out and 16 feet feet above her starting point and the floor is 8 feet below her starting point. Use the focus of (8,16) and the directrix at y= -8 to determine the equation of the parabola.

Answer 1

Recall that the focus of a parabola is a point inside the parabola along the axis of symmetry of the parabola with the same distance from the vertex of the parabola as the directrix of the parabola.

Given that the focus of the parabola is (8, 16) i.e. a y-value of 16. The vertex of a parabola is halfway the distance between the y-values of the focus and the directrix.
i.e. the vertex of the parabola will have a y-value of (16 + (-8)) / 2 = (16 - 8) / 2 = 8 / 2 = 4.
Thus the vertex of the parabola is (8, 4)

Let point (x, y) be any point on the parabola, the distance between point (x, y) and the focus is
$\sqrt{(x-8)^2+(y-16)^2}$
while the distance between the point (x, y) and the directrix is
$|y-(-8)|=|y+8|$

Now, by definition, the distance between any point in a parabola and the focus is equal to the distance between that point and the directrix.
i.e.
$\sqrt{(x-8)^2+(y-16)^2}=|y+8| \\ (x-8)^2+(y-16)^2=(y+8)^2 \\ x^2-16x+64+y^2-32y+256=y^2+16y+64 \\ 16y+32y=x^2-16x+256 \\ 48y=x^2-16x+256 \\ y= \frac{1}{48} x^2- \frac{1}{3} x+ \frac{16}{3}$