Change to standard form of conic PLz help

[aisha]

Write y^2-4x-4y+8=0 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 im not sure how to factor it should have been in the form ax^2 + by^2 + 2gx + 2fy + c = 0 but the x^2 term is missing, so i dont 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 plz help me with the first step how to get this in to standard form.

[Data]

What type of conic has only one of the variables squared in its standard form?

[aisha]

yeh thats what im 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 cuz 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
(x^2 - 4x)+ (y^2 -4y) =-8
(x^2 -4x +4) + (y^2 -4y +4)=-8+4+4
(x-2)^2 + (y-2)^2 = 0

What is the problem?

[Data]

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).

[aisha]

opps lol sorry

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

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

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

I dont know how to get the equation given into standard form at all plz help me, I have tried with ellipse but never parabola.

[Data]

Solve for x in terms of y, then complete the square for the y terms.

[aisha]

Im not sure I understand what you mean.

[Data]

y^2-4x-4y+8=0 \Longrightarrow 4x = y^2 - 4y + 8

then complete the square on the right side.

[aisha]

when I completed the square on the RHS I got

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

what do I do now?

[Data]

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.

[aisha]

is that +4 supposed to stay on that side? How come this equation doesnt 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:

[Data]

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

[aisha]

Does P=1 what are the points for the focus?

[seiferseph]

when I completed the square on the RHS I got

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

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

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

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