Question

Find the equation for a parabola with a directrix of x=5, and focus of (3, 8).

Answer 1

(y-8)² = -4 (x-4)

Explanation:

Directrix x=5 so we have a horizontal parabola

In general directrix for a horizontal parabola is x= h -p

Focus is (3, 8)

In general focus for a horizontal parabola is (p+h, k)

From this we can conclude that:

k=8

h -p =5

h+p = 3

We have two equation with two unknown. We can solve and find h and p.

h -p =5 → h = p+5

h+p = 3 → p+5+p = 3 → 2p = 3-5 → p = -2/2 → p = -1

h = p+5 → h = -1+5 → h = 4

Standard equation of a horizontal parabola is

(y - k)² = 4p (x - h), substitute k= 8, p= -1, and h=4

(y - 8)² = 4(-1) (x - 4)

(y - 8)² = -4 (x - 4)

Answer 2

`(y-8)^2=-4(x-4)`

Explanation:

Because the directrix is a vertical line, we must use the equation
`(y-k)^2=4p(x-h)` where
`p\\eq 0`. The vertex of this parabola is at
`(h,k)`, the focus point is at
`(h+p,k)`, and the directrix is at the line
`x=h-p`.

Since the directrix is at the line
`x=5`, then we have the equation
`5=h-p`.

Because the focus point is at
`(3,8)`, then the equation
`3=h+p` will help determine the x-coordinate of the focus.

Now, we solve the system of equations:

`\displaystyle \left \{ {{5=h-p} \atop {3=h+p}} \right.\\\\2=-2p\\-1=p`

`5=h-p\\\\5=h-(-1)\\\\5=h+1\\\\4=h`

Thus, the equation for the parabola will be
`(y-8)^2=4(-1)(x-4)\rightarrow(y-8)^2=-4(x-4)` with vertex
`(h,k)\rightarrow(4,8)`.