How do you find the vertex, focus and directrix of x^2-2x+8y+9=0x2−2x+8y+9=0?

1 Answer
Shell
Dec 16, 2016

The vertex is (1,-1)(1,−1), the focus is (1,-3)(1,−3), and the directrix is y=1y=1.

Explanation:

x^2-2x+8y+9=0x2−2x+8y+9=0

Complete the square to put the equation in the vertex form of a parabola.

x^2-2xcolor(white)(aaa)=-8y-9x2−2xaaa=−8y−9

Divide the coefficient of the xx term by 2 and square it. Then add the square to both sides of the equation.

-2/2 =-1 => (-1)^2=1−22=−1⇒(−1)2=1

x^2-2x+1=-8y-9+1x2−2x+1=−8y−9+1

Factor the left side. Note that the constant term in the factors is the quotient of the coefficient of the xx term divided by 2.

(x-1)(x-1)=-8y-8(x−1)(x−1)=−8y−8

Rewrite the left side as the square of a binomial, and factor out a GCF from the right side.

(x-1)^2=-8(y+1)(x−1)2=−8(y+1)

Rearrange to match the vertex form of a parabola that opens up or down: 4p(y-k)=(x-h)^24p(y−k)=(x−h)2 where (h,k) (h,k) is the vertex and pp is used to find the focus and directrix, as shown below.

-8(y+1)=(x-1)^2−8(y+1)=(x−1)2

The vertex (h,k)=(1,-1)(h,k)=(1,−1)

Because the 4p=-84p=−8 is negative, the parabola opens down.

p=-2p=−2

The focus is "inside" the parabola, pp units away from the vertex. Because the parabola opens down, the focus is BELOW the vertex, and its coordinates are (1,-3)(1,−3)

The directrix is a line "outside" the parabola, and wil be a horizontal line two units ABOVE the vertex. The equation of the directrix is y=1y=1.

Related questions
See all questions in Identify Critical Points
Impact of this question
3651 views around the world
You can reuse this answer
Creative Commons License