How do you find the vertex, focus and directrix of x^2-2x+8y+9=0x22x+8y+9=0?

1 Answer
Feb 4, 2016

The vertex is at (1,-1)(1,1),
The focus is (1,-5)(1,5) and the directrix is at y=3y=3

Explanation:

The standard form of a parabola is written as y = ax^2 +bx +cy=ax2+bx+c
or in vertex form as y =a(x-h)^2 +ky=a(xh)2+k where (h,k)(h,k) is the vertex and 1/(2a)12a is the distance between the vertex and the focus as well the distance between the vertex and the directrix.

Firstly rearrange the expression to get yy alone on the left had side.

x^2-2x+8y+9 = 0x22x+8y+9=0

8y = -x^2 +2x -98y=x2+2x9

y = -1/8(x^2 - 2x +9)y=18(x22x+9)

Now put it into vertex form by completing the square

y = -1/8((x-1)^2 -1+9) = -1/8((x-1)^2+8)y=18((x1)21+9)=18((x1)2+8)

y = -1/8(x^2-1)^2 -1y=18(x21)21

The vertex is at (1,-1)(1,1).

Because aa is negative the vertex is the maximum value of the equation - all values of xx give values less than the vertex

a=-1/8a=18 so 1/(2a) = 1/(-2/8) =-412a=128=4

The focus is therefore (1,-5)(1,5) and the directrix is at y=3y=3