Question

What are the coordinates of the focus of the parabola? (x+1)2=−8(y−2) (−1, 2) (1, −2) (−1, 0) (−1, 1)

Answer 1

`(x,y) = (-1,0)`

Explanation:

Given

`(x+1)^2 = -8(y-2)`

Required

The coordinates of the focus

First, write the expression in form of:
`y=a(x-h)^2+k`

`(x+1)^2 = -8(y-2)`

`(x+1)^2 = -8y + 16`

`-8y + 16 = (x+1)^2`

Subtract 16 from both sides

`-8y = (x+1)^2 - 16`

Divide through by -8

`y = -(1)/(8)(x+1)^2 - (16)/(-8)`

`y = -(1)/(8)(x+1)^2 +2`

In this case:

`a = -(1)/(8)`

`-h = 1`
`h = -1`

`k = 2`

The focus of the parabola is:

`(x,y) = (h,k + (1)/(4a))`

This gives:

`(x,y) = (-1,2 + (1)/(4*-(1)/(8)))`

`(x,y) = (-1,2 + (1)/(-(1)/(2)))`

`(x,y) = (-1,2 -2)`

`(x,y) = (-1,0)`