Question

What is the equation of the quadratic graph with a focus of (3, 1) and a directrix of y = 5?

Answer 1

Answer:

$f(x)=-\frac{1}{8}(x-3)^2+3$

Step-by-step explanation:

We want to find the equation of the parabola with a focus of $(3,1)$ and directrix $y=5$.

Considering the directrix, the quadratic graph must open downwards.

The equation of this parabola is given by the formula,

$(x-h)^2=4p(y-k)$, where $(h,k)$ is the vertex of the parabola.

The axis of this parabola meets the directrix at $(3,5)$.

Since the vertex is the midpoint of the focus and the point of intersection of the axis of the parabola and the directrix,

$h=\frac{3+3}{2}=3$ and $k=\frac{5+1}{2}=3$.

The equation of the parabola now becomes,

$(x-3)^2=4p(y-3)$.

Also $|p|$ is the distance between the vertex and the directrix.

$|p|=2$

This implies that $p=-2\:or\:2$.

Since the parabola turns downwards,

$p=-2$.

Our equation now becomes,

$(x-3)^2=4(-2)(y-3)$.

$(x-3)^2=-8(y-3)$.

We make y the subject to get,

$y=-\frac{1}{8}(x-3)^2+3)$.

This is the same as

$f(x)=-\frac{1}{8}(x-3)^2+3)$.