How Do You Convert the Equation y^2-4x-4y+8=0 into Standard Form?

Click For Summary

Homework Help Overview

The discussion revolves around converting the equation y² - 4x - 4y + 8 = 0 into standard form and identifying its key features, particularly focusing on the type of conic represented by the equation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to complete the square to convert the equation into standard form. There is uncertainty about how to handle the absence of an x² term and whether it indicates a typo. Some participants explore the characteristics of conics with only one variable squared, while others attempt to manipulate the equation to identify its features.

Discussion Status

Several participants have offered guidance on completing the square and isolating variables. There is ongoing exploration of the implications of the equation's structure, with some participants questioning the nature of the conic and its standard form representation. Multiple interpretations of the equation's characteristics are being examined.

Contextual Notes

Participants express confusion regarding the initial equation's format and its implications for identifying the conic type. There are references to specific forms of parabolas and the need for clarity on focal lengths and key features, indicating a focus on understanding rather than resolving the problem.

aisha
Messages
584
Reaction score
0
Write [tex]y^2-4x-4y+8=0[/tex] in standard form and state its key features, Identify the type of conic.

I know to get into standard form I need to complete the square but with these numbers I am not sure how to factor it should have been in the form [tex]ax^2 + by^2 + 2gx + 2fy + c = 0[/tex] but the x^2 term is missing, so i don't know what to group in brackets and then what to factor.

I think after I get this question into standard form I will be able to tell what type of conic it is and state the key features please help me with the first step how to get this into standard form.
 
Last edited:
Physics news on Phys.org
What type of conic has only one of the variables squared in its standard form?
 
yeh that's what I am thinking I think it is a MAJOR typo, you know what? I think they forgot to add an x^2

If there is an x^2 then this is the equation of a circle right because the values of a and b are equal?

this is what I did but somethin is wrong because I got the radius as 0

after adding x^2 to the eqn I grouped the x and y terms and took the 8 to the RHS
[tex](x^2 - 4x)+ (y^2 -4y) =-8[/tex]
[tex](x^2 -4x +4) + (y^2 -4y +4)=-8+4+4[/tex]
[tex](x-2)^2 + (y-2)^2 = 0[/tex]

What is the problem?
 
It's probably not a typo. Like I asked, can you think of a type of conic that has only one of the variables squared in its standard form (I was being serious! :smile:)? Review all the types of conics if you need to. There are four (well, really three, but you are probably counting circles separately from ellipses).
 
opps lol sorry

[tex](x - h)^2 = 4p(y - k)[/tex] a parabola opening up or down

[tex](y - k)^2 = 4p(x - h)[/tex] a parabola opening left or right

these two equations for parabolas in standard form only require one term to be squared.

I don't know how to get the equation given into standard form at all please help me, I have tried with ellipse but never parabola.
 
Last edited:
Solve for [itex]x[/itex] in terms of [itex]y[/itex], then complete the square for the [itex]y[/itex] terms.
 
Im not sure I understand what you mean.
 
[tex]y^2-4x-4y+8=0 \Longrightarrow 4x = y^2 - 4y + 8[/tex]

then complete the square on the right side.
 
when I completed the square on the RHS I got

[tex]4x= (y-2)^2 +4[/tex]

what do I do now?
 
Last edited:
  • #10
That's fine. Now you just need to identify all the important characteristics and such. Is it opening left/right or up/down? Where is its vertex? etc.
 
  • #11
is that +4 supposed to stay on that side? How come this equation doesn't really look like one of the parabola standard form equations?

I think the center is (2,4)

How do you figure out the value of P (focal length)?

and the point where the focus is and the equation of the direct x?

The reason I am asking is because the example I have is on this equation
State the key feature of (x -1)^2 = 12 (y -2) Is my equation like this one? Can you put my equation in this form so that I can follow the example to find all of the key features? :redface:
 
  • #12
[tex]4x = (y-2)^2+4 \Longrightarrow 4(x-1) = (y-2)^2[/tex]
 
  • #13
Does P=1 what are the points for the focus?
 
  • #14
aisha said:
when I completed the square on the RHS I got

[tex]4x= (y-2)^2 +4[/tex]

what do I do now?

i've just done stuff like this now

you divide by 4 to isolate x, and you get an inverse of a parabola

[tex]x= 1/4(y-2)^2 +1[/tex]

you can easily get the vertex and axis of symmetry from this, as well as y and x interecepts
 

Similar threads

Replies
7
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
6
Views
2K