Write an equation of a parabola with a vertex at the origin and a directrix at y = 5.
AAAaannndd the bot gets it wrong yet again!
y=0 is not even a parabola!
recall that x^2 = 4py has vertex at (0,0) and directrix y=-p
so we have p = -5, giving us
x^2 = -20y
y = -1/20 x^2
you can check this at
wolframalpha. com/input?i=parabola+y+%3D+-1%2F20+x%5E2
Since the vertex is at the origin, the equation of the parabola takes the form y = ax^2.
The distance from a point (x,y) on the parabola to the directrix y = 5 is |y - 5|. The distance from the same point to the vertex (0,0) is equal to the distance from the point to the focus (0,-a/4).
Using the distance formula, we have:
|y - 5| = dist((x,y), (0,-a/4))
= sqrt((x-0)^2 + (y + a/4)^2)
= sqrt(x^2 + (y + a/4)^2)
Squaring both sides, we get:
(y - 5)^2 = x^2 + (y + a/4)^2
Expanding and simplifying, we get:
x^2 + (3/16)a^2 - 5y = 0
We know that the vertex is at (0,0), so plugging this in gives:
0 + (3/16)a^2 - 5(0) = 0
Simplifying, we get:
a^2 = 0
Therefore, the equation of the parabola is y = 0.