Question

What is the equation of the quadratic graph with a focus of (6, 0) and a directrix of y = −10 ?

Answer 1

(x-6)^2 + y^2 = (y +10)^2 (x-6)^2 = (y +10)^2 - y^2 = (y+10+y)(y+10-y) = (2y+10)(10) = 2(y+5)(10) = 20(y+5) 1/20*(x-6)^2 = y+5 y = 1/20*(x-6)^2 - 5

:)

Answer 2

`(x-6)^(2)=20(y+5)^(2)`

Explanation:

The standard for for the equation of a parabola is
`(x-h)^(2)=4p(y-k)`

The focus is given as
`(h, k+p)`

The directrix is given by
`y=k-p`

• Comparing focus given as
  `(6,0)` to formula of focus
  `(h,k+p)` , we see that
  `h=6` and
  `k+p=0`
• Comparing directrix given as
  `y=-10` to formula of directrix
  `y=k-p` , we can write
  `k-p=-10`

We can use the two system of equations [
`k+p=0` and
`k-p=-10` ] to solve for
`k` and
`p`.

Adding the two equations gives us:

`2k=-10\\k=(-10)/(2)\\k=-5`

Using this value of
`k` , we can find
`p` by plugging this value in either equation. Let's put it in Equation 1. We have:

`k-p=-10\\-5-p=-10\\-5+10=p\\p=5`

Now, that we know
`h=6` ,
`k=-5` , and
`p=5` , we can plug these values in the standard form equation to figure out the parabola's equation. Doing so and rearranging gives us:

`(x-6)^(2)=4(5)(y-(-5))^(2)\\(x-6)^(2)=20(y+5)^(2)\\`

This is the equation of the parabola with focus
`(6,0)` and directrix
`y=-10`