Question

The parabola described by the equation -1/16(x-3)^2 + 4 = y has the directrix at y=8. What is the focus of the parabola?

Answer 1

(3, 0)

Explanation:

A parabola is the locus of a point such that the distance from a fixed line called directrix and the distance from a fixed point called focus is constant.

The equation of a parabola with a vertex at (h, k) with axis of symmetry parallel to the y axis is given as:

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

The directrix is at y = k - p, and focus is at (h, k + p).

Given an equation -1/16(x-3)^2 + 4 = y, expressing it in standard form is:

`-(1)/(16)(x-3)^2+4=y\\\\(y-4)= -(1)/(16)(x-3)^2\\\\-16(y-4)=(x-3)^2\\\\`

Comparing with the standard form:

The center = (h, k) = (3, 4)

Also, -16 = 4p

p = -4

Directrix is at y = k - p = 4 - (-4) = 8

Directrix is at y = 8

The focus is at (h, k + p) = (3, 4 + (-4)) = (3, 0)

The focus is at (3, 0)