Need help with Locus (1 Viewer)

[Tabris]

4y = x^2-4x+8

find vertex
find directrix

i got stuck on this when doing revision

is this right?

4ay = x^2
therefore a = 1
since this type of parabola has its vertex at 0,0 then the vertex is 0,1 and directrix y = -1?

 

[CM_Tutor]

The form you need is (x - h)² = 4a(y - k), where a is the focal length and (h, k) is the vertex, which is the general form for when the vertex is not at the origin (which it can't be here, as the origin does not lie on the parabola).

In this case: 4y = x² - 4x + 8
4y - 8 + (-4 / 2)² = x² - 4x + (-4 / 2)²
4y - 4 = x² - 4x + 4
4(y - 1) = (x - 2)²

So, the vertex is at (2, 1), and the focal length is 1 (as 4a = 4).
So, the focus is at (2, 2), and the directrix is y = 0

 

[Tabris]

I get it now

u subtracted 8 from both side and added 4 on both sides to complete the square, then u leg LHS = 4(y-1) and RHS (x-2)^2

therefore the vertex is (2,1) and since 4a(y-1) focal length a = 1 therefore he directrix is y = o

thx alot champ