Write an equation of the parabola with an axis of symmetry of x=0 and whose focus is 8 units above the directrix y=−10 .

Answer 1

The basic equations are:

(x - h)² = 4p (y - k) and (y - k)² = 4p (x - h)

When I sketch the directrix and the focus, this shows that the parabola opens up. This means it would use this version of the equation.

(x - h)² = 4p (y - k)

The vertex is (h,k). It will be exaxtly half way between the focus and the directrix, this would be (0,-1) The p value is the distance between the vertex and the focus and the vertex and the directrix. In this case, 9. Plugging all this in.

(x - 0)² = 4(9) (y - (-1)) Simplify

x² = 35(y+1)

Answer 2

A parabola with a directrix y = (some number) means it is a parabola going up and down like this:

In other words it is in the form y = x² (as opposed to x = y²)

If the focus is above the directrix then the parabola opens upward (smiley face) and has a positive stretch factor (as shown above).

The vertex is halfway between the directrix and the focus. The vertex also is in line with the focus (or you could say that the line of symmetry goes through both the vertex and the focus). If the vertex is halfway between the directrix and the focus, then it must have a y value of y = -6 and, since it is in line with the axis of symmetry, the vertex has an x value of 0.

Using the vertex form of a quadratic, y = a(x - 0)² - 6 or, simplifying, y = ax² - 6. The value of "a" is determined by the distance from the vertex to the focus or the the directrix. It is 4 units away from both so the "p" value is 4 and "a" is defined by a = 1/(4p) so a = 1/(4*4) = 1/16.

So the equation is y = 1/16x² - 6