Locus Help (1 Viewer)

[Shoom]

A parabola has equation 4y= x^2+4x+8

How can I rearrange it in the form (x-h)^2=4a(y-k) can please show the steps.

I also need to know how to complete the sqaures and in what situations in locus will I need to use complete the square ( what type of question)

thanks.

 

[omniscience]

A parabola has equation 4y= x^2+4x+8

How can I rearrange it in the form (x-h)^2=4a(y-k) can please show the steps.

I also need to know how to complete the sqaures and in what situations in locus will I need to use complete the square ( what type of question)

thanks.

ok.

4y = x^2 + 4x + 8

Using the "completing the square rule", you can get:

4y - 4 = x^2 + 4x + 4
4(y-1) = (x+2)^2

To "complete the square", you first divide the value of x (which is 4 here) by two and square it (and you get 4 from that) and that's pretty much it. So you need 4 to "complete the square". I achieved that by taking 4 out from both sides.

 

[iluvGG]

hi

i think i know how to do this question, but anyone feel free to correct me.

you need to complete the square on the RHS of the original equation.

so you get: 4y+4 = x^2+4x+4+8 (simplify x^2+4x+4 into perfect square form)
4y+4 = (x + 2)^2 + 8 (carry 8 over to the other side)
4y-4 = (x+2)^2
(x+2)^2 = 4(y-1) --> required form

hope that is right and is understandable!!! sorry if its not. . .

 

[Shoom]

So whatever is in front of the X value and only the x value I divide that by 2 and square it and then I subtract that number from both sides of the equation?

So for the equation y=2x^2+4x-1 when I complete the square it should become

y-4=2x^2+4x-5?

 

[omniscience]

So whatever is in front of the X value and only the x value I divide that by 2 and square it and then I subtract that number from both sides of the equation?

Well, yeah.

The idea is to get all desired value to make the "complete equation". I hope that cleared some up.

 

[Shoom]

2x^2+4x-5

Whats its factorised form?

Thanks for all the help
wait because this is a locus do I have to something different I need to get the full equation into the same form again

 

[omniscience]

2x^2+4x-5

Whats its factorised form?

Thanks for all the help

Factorised form?

I would personally use Quadratic formula to find the roots.

If you didn't mean factorised form but complete equation,

I would first go: 2x^2 + 4x - 5 = 2(x^2 + 2x -5/2)
= 2(x^2 + 2x + 1 -7/2)
= 2(x^2 + 2x + 1) - 7
= 2(x+1)^2 - 7

 

[Shoom]

In my solution from the school, the second example I posted they complete the square by using the 2 infront of the X^2 is that because they take it from the higest power? wth they add the complete sqaure aswell so it look s like this

y+1= 2x^2 + 4x
1/2(y+1)=x^2+2x
1/2y+1/2+1=x^2+2x+1
(x+1)^2= 4 ( 1/8) ( y+3)

that example lost me

 

[omniscience]

In my solution from the school, the second example I posted they complete the square by using the 2 infront of the X^2 is that because they take it from the higest power?

Whatever method you use, if it works, use it. I used that method because that was easy for me.

 

[Shoom]

I tried to use yours the only one I know of lol. But the solution posts something strange, do you follow the solution I gave for the second example? I dont. thats from the exam marker.

 

[omniscience]

I tried to use yours the only one I know of lol. But the solution posts something strange, do you follow the solution I gave for the second example? I dont. thats from the exam marker.

Ok, first, you didn't give me the y +1 in the equation.

I will re-do it for you, wait a minute.

 

[omniscience]

y+1= 2x^2 + 4x
1/2(y+1)=x^2+2x
1/2y+1/2+1=x^2+2x+1
(x+1)^2= 1/2(y+3)

 

[Shoom]

Last question I promise.
The original question is y=2x^2+4x-1
how do you get from y=2x^2+4x-1 to y+1= 2x^2 + 4x

 

[omniscience]

Last question I promise.
The original question is y=2x^2+4x-1
how do you get from y=2x^2+4x-1 to y+1= 2x^2 + 4x

I added 1 to both sides.

 

[Shoom]

yeah I know, any particular reason?

 

[omniscience]

yeah I know, any particular reason?

I didn't initially do that. It was you who provided the working out.