State the vertex, focus and directrix of each parabola

Question

State the vertex, focus and directrix of each parabola

y^2 - 8y + 8x = 0

Answer 1

Rewrite the quadratic equation in a standard parabola form:

x-h = a(y-k)^2

x = m(y-k)^2 + h

8x = -y^2+8y +16 -16

x = (-1/8)(y+4)^2 -2 = 1/4a(y-k)^2 + h

the -1/8 means the parabola opens to the left

the vertex is (h,k) = (2,4)

the y-intercepts are the points where the parabola intersects the y-axis

set x=0, solve for y

y^2 - 8y +8(0) = 0

y^2 -8y = 0

y(y-8) = 0

y=0 and 8

the vertex has a y coordinate in the middle of 0 and 8, or 4. The focus also has the same y coordinate.

take the coefficient of the squared y term and set it equal to 1/4a. If this were an up or downward opening parabola, you'd set the coefficient equal to 4a and solve for a

But here 1/4a = 1/8

and 4a = 8

a = 2

2 is the distance from the vertex to the focus

and the distance from the vertex to the directrix line

the Vertex (2,4) is 2 from the focus on the left (0,4)

and the vertex is 2 from the directrix on the right, x=4

directrix: x=4

focus (0,4)

the focus

Answer 2

Because it is the "y" that is squared we know that the parabola has a horizontal axes of symmetry. The standard form will look like this: x=a(y-k)²+k where the vertex = (h,k), focus = (h+(1/4a),k) and directrix is x = h - (1/4a).

Let's rearrange our equation to get it with x=....: 8x=-y²+8y

8x=(-1)(y²-8y) "pull out" a negative 1

8x=(-1)(y²-8y+16)+16 complete the square

8x=(-1)(y-4)²+16 simplify

x=(-1/8)(y-4)²+2 divide both sides by 8

We now have our equation in standard form. All we need to do now is determine a, h and k and put them into their proper places for the vertex, focus and directrix. We can see from the equation that a=-1/8, h=2 and k=4. Therefore, vertex=(h,k)=(2,4), focus=(h+(1/4a),k)=(2+(1/(4)(-1/8)),4)=(2+(1/(-1/2)),4)=(2-2,4)=(0,4) and directrix x=h-(1/4a)--->x=2-(1/(4)(-1/8))--->x=2+2--->x=4.

Hope this helps!