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

Answer:

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

Step-by-step 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

$y = -\frac{1}{8}(x+1)^2 - \frac{16}{-8}$

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

In this case:

$a = -\frac{1}{8}$

$-h = 1$         $h = -1$

$k = 2$

The focus of the parabola is:

$(x,y) = (h,k + \frac{1}{4a})$

This gives:

$(x,y) = (-1,2 + \frac{1}{4*-\frac{1}{8}})$

$(x,y) = (-1,2 + \frac{1}{-\frac{1}{2}})$

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

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