Parabola (Horizontal) - finding vertex, focus and equation of line

1. Oct 3, 2009

zebra1707

1. The problem statement, all variables and given/known data

Use completing the square method to rewrite the equation of the parabola
y^2 – 4y – 44 = 16x in the form (y-y0)^2 = 4A(x-x0)

Hence find:

a) the coordinates of the vertex
b) the coordinates of the focus
c) the equation of the line that passes through the focus and parallel to the y-axis.

y^2 - 4y - 44 = 16x at this point I thought that the only way that I could get an equation whereby I could complete the square (and in the form required) was to add 48 to both sides of the equation.

which would give me y^2 - 4y + 4 = 16x + 48 and I could complete the square and in the form required.

(y - 2)^2 = 16(x+3)

*Vertex therefore would be (2, -3)
*Focal length A = 4
*Focus S is 4 units to the right of the Vertex S(6, -3)
*Equation of the line that passes through the focus and parallel to the y-axis x = -3

2. Relevant equations

As above

3. The attempt at a solution

As above - can someone confirm that I am on the right track with this?

2. Oct 3, 2009

symbolipoint

The square piece which seems to be needed for adjustment purposes is (-4/2)^2, so you would add and subtract +4.

One of your steps should include something of lefthand side as
(y^2 - 4y + 4) - 4 - 44 which you should find by inspection contains a factorable expression, ultimately allowing you to find x as a function of y, in standard form ( I did not show all of the steps or full equations).

3. Oct 3, 2009

zebra1707

Hi there - many thanks for the assistance.

Just so that I have this right.

y^2 - 4y - 44 = 16x

y^2 - 4y + 4 - 44 = 16x + 4

(y^2 - 4y + 4) - 44 = 16x + 4

(y-2)^2 = 16x + 4 + 44

(y-2)^2 = 16x + 48

(y-2)^2 = 16(x + 3) I dont think that I need to simplify this any further?

Cheers Petra

4. Oct 3, 2009

zebra1707

Are these correct?

*Vertex therefore would be (2, -3) - No (-3, 2)
*Focal length A = 4
*Focus S is 4 units to the right of the Vertex S( )
*Equation of the line that passes through the focus and parallel to the y-axis x = -3

5. Oct 3, 2009

zebra1707

Are these correct?

*Vertex therefore would be (2, -3) - No (-3, 2)
*Focal length A = 4
*Focus S is 4 units to the right of the Vertex S(1,2)
*Equation of the line that passes through the focus and parallel to the y-axis x = 1

6. Oct 3, 2009

symbolipoint

Your algebra steps seem good, but you did not finish. Your first goal was determine how to complete the square. Your second goal is to find x as a function of y and have this be in standard form. Reading the vertex from the result would then be rendered easily.

7. Oct 3, 2009

zebra1707

Hi there

The question asks to "rewrite the equation" of the parabola in the form (y-yo)^2 = 4A(x-xo)
and from there find the vertex etc.... From the point that I have reached I can Identify the vertex.

So Im not sure that I understand you second statement? Just a bit confused.

Cheers Petra d