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

In summary, to write the equation y^2-4x-4y+8=0 in standard form, we need to complete the square and rearrange the terms to get it in the form ax^2 + by^2 + 2gx + 2fy + c = 0. After completing the square, we can identify the type of conic by looking at the coefficients of x and y. We can also determine the key features, such as the vertex, focus, and axis of symmetry, by using the standard form equation.
  • #1
aisha
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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.
 
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  • #2
What type of conic has only one of the variables squared in its standard form?
 
  • #3
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?
 
  • #4
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).
 
  • #5
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.
 
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  • #6
Solve for [itex]x[/itex] in terms of [itex]y[/itex], then complete the square for the [itex]y[/itex] terms.
 
  • #7
Im not sure I understand what you mean.
 
  • #8
[tex] y^2-4x-4y+8=0 \Longrightarrow 4x = y^2 - 4y + 8[/tex]

then complete the square on the right side.
 
  • #9
when I completed the square on the RHS I got

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

what do I do now?
 
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  • #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
 

1. What is standard form for conic equations?

Standard form for conic equations is an algebraic expression that represents a conic section in the form Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0, where A, B, and C are not all zero and x and y are variables. This form is used to easily identify the type of conic section and its key properties.

2. How do you convert a conic equation to standard form?

To convert a conic equation to standard form, you can use a process called completing the square. This involves manipulating the equation to group the x^2 and y^2 terms together, and then adding or subtracting a constant to make the squared terms perfect squares. The resulting equation will be in standard form.

3. Why is standard form important in studying conic sections?

Standard form is important in studying conic sections because it allows us to easily identify the type of conic section and its key properties. For example, if the coefficient of the x^2 term is positive, it represents an ellipse, and if it is negative, it represents a hyperbola. This form also helps in graphing and solving equations involving conic sections.

4. Can all conic equations be converted to standard form?

Yes, all conic equations can be converted to standard form. However, some equations may require more complex algebraic techniques to manipulate and convert to this form. It is also important to note that some equations may not have real solutions in standard form, indicating that they do not represent a conic section.

5. Are there any other forms for representing conic equations?

Yes, there are other forms for representing conic equations, such as general form, vertex form, and polar form. However, standard form is the most commonly used form for studying and working with conic sections.

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