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

[zebra1707]

Homework Statement

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

Homework Equations

As above

The Attempt at a Solution

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

 

[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).

 

[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 don't think that I need to simplify this any further?


Cheers Petra

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).

 

[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

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 don't think that I need to simplify this any further?


Cheers Petra

 

[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

 

[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.

 

[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 I am not sure that I understand you second statement? Just a bit confused.

Cheers Petra d

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.