Parabolas as conic sections

**Hapa** · April 26th 2009, 07:43 PM

Still not quite sure if I am performing these right.

1. x - 4 = (y + 2)^2
x - 4 = y^2 + 4y + 4
x = y^2 + 4y + 8 (after this step, i am unsure if i am completing the square correctly)

does the step above equal to
x = (y^2 + 4y + 4) +8 + 4
x = (y + 2)^2 + 12

this would make the vertex (12, -2)? for some reason, this doesn't make sense. and then to find the "y" intercepts, i would just plug into the quadratic formula, correct?

a simpler one that is throwing me off is

x = 2y^2 + 3

do i complete the square with this one as well? I know that since you have to get this into a the form "x = a(y - k)^2 + h" the 2 signifies that this parabola will open up to the right. but as far as figuring out the vertex from only 2 terms, i am getting mixed up.

the last problem i have is the following

x = -y^2 + 4y - 3

since there is a negative sign in front of the y squared, do i go ahead and factor out a (-1), complete the square, and then plug into the quadratic formula? thanks again for everyone's help! hope you all had a wonderful weekend!

hapa

**earboth** · April 27th 2009, 01:33 AM

Originally Posted by Hapa

Still not quite sure if I am performing these right.

1. x - 4 = (y + 2)^2
x - 4 = y^2 + 4y + 4
x = y^2 + 4y + 8 (after this step, i am unsure if i am completing the square correctly)

does the step above equal to
x = (y^2 + 4y + 4) +8 + 4
x = (y + 2)^2 + 12

this would make the vertex (12, -2)? for some reason, this doesn't make sense. and then to find the "y" intercepts, i would just plug into the quadratic formula, correct?

It would be of some help if you could tell us what you want to calculate. If you are looking for the coordinates of the vertex then it isn't necessary to transform the given equation. You can pick up the coordinates as V(4, -2)

a simpler one that is throwing me off is

x = 2y^2 + 3

do i complete the square with this one as well? I know that since you have to get this into a the form "x = a(y - k)^2 + h" the 2 signifies that this parabola will open up to the right. but as far as figuring out the vertex from only 2 terms, i am getting mixed up.

If the vertext has the coordinates $V(x_V\ ,\ y_V)$ then the general equation of a parabola opening right is:

$(x-x_V)=4\cdot p \cdot (y-y_V)^2$

Now transform the given equation into this form:

$x = 2y^2+3~\implies~x-3=4 \cdot \dfrac12 \cdot (y-0)^2$

Therefore the vertex is at V(3, 0)

the last problem i have is the following

x = -y^2 + 4y - 3

since there is a negative sign in front of the y squared, do i go ahead and factor out a (-1), complete the square, and then plug into the quadratic formula? thanks again for everyone's help! hope you all had a wonderful weekend!

hapa

You are dealing with a parabola opening left:

$x = -y^2 + 4y - 3~\implies~ x+3 \bold{{\color{red}- 4}} = -(y^2-4y\bold{{\color{red}+ 4}})$ will yield:

$x-1 = -4\cdot \dfrac14 \cdot (y-2)^2$

**HallsofIvy** · April 27th 2009, 03:09 AM

Originally Posted by Hapa

Still not quite sure if I am performing these right.

1. x - 4 = (y + 2)^2
x - 4 = y^2 + 4y + 4
x = y^2 + 4y + 8 (after this step, i am unsure if i am completing the square correctly)

The square was already "completed"! x= (y+ 2)^2+ 4. The vertex is at (4, -2) and the axis of symmetry is horizontal.

does the step above equal to
x = (y^2 + 4y + 4) +8 + 4

No, it is equal to x= (y^2+ 4y+ 4)+ 8 -4 which gives back exactly what you started with.

x = (y + 2)^2 + 12

this would make the vertex (12, -2)? for some reason, this doesn't make sense. and then to find the "y" intercepts, i would just plug into the quadratic formula, correct?

a simpler one that is throwing me off is

x = 2y^2 + 3

do i complete the square with this one as well?

Do you understand what "completing the square" means. Since there is no "y to the first power" in this, just as in the previous one, it is already a "perfect square". The vertex is at (3, 0).

I know that since you have to get this into a the form "x = a(y - k)^2 + h" the 2 signifies that this parabola will open up to the right. but as far as figuring out the vertex from only 2 terms, i am getting mixed up.

Again, it is already in that form: with a= 2, k= 0, and h= 3.

the last problem i have is the following

x = -y^2 + 4y - 3

since there is a negative sign in front of the y squared, do i go ahead and factor out a (-1), complete the square, and then plug into the quadratic formula? thanks again for everyone's help! hope you all had a wonderful weekend!

hapa

Yes. x= -(y^2- 4y)- 3. Since 4/2= 2 and 2^2= 4, that is the same as
x= -(y^2- 4y+ 4- 4)- 3= -(y^2- 4y+ 4)+ 4- 3= -(y+ 2)^2+ 1.

**Soroban** · April 27th 2009, 03:29 AM

Hello, Hapa!

You're making me dizzy . . .

$(1)\;\;x - 4 \:=\: (y + 2)^2$

Why are you multiplying it out ... then trying to complete the square?

Do you want the vertex? . . . It's at $(4,-2)$ . . . right?

Do you want the y-intercepts? . . . Let $x = 0$ and solve for $y.$

$x \:= \:2y^2 + 3$

Do you want the form: . $x \:=\:a(y-k)^2 + h$ ?

You've got it! . . . $x \:=\:2(y-0)^2 + 3$

$x \:= \:-y^2 + 4y - 3$

Multiply both sides by -1: . $-x \;=\;y^2 - 4x + 3$

Complete the square: . $-x \:=\:y^2 - 4x \:{\color{red}\:+\: 4} + 3 \:{\color{red}-\:4}$

. . . . . . . . . . . . . . . . $-x \:=\:(y-2)^2 - 1$

Multiply by -1: . . . . . . $x \:=\:-(y-2)^2 + 1$

Thread: Parabolas as conic sections

LinkBack

Thread Tools

Display

Parabolas as conic sections

Similar Math Help Forum Discussions

Conic Sections

Conic Sections

Conic sections

Algebra 2 Conic Sections - Parabolas

conic sections

Search Tags