How do you find the vertex, focus, and directric of the parabola given by the equation :
x^2 + 4x + 4y + 16 = 0
2006-06-26 06:05:30 · 4 answers · asked by amber j 1 in Science & Mathematics ➔ Mathematics
Solve for y, and complete the square for x.
4y = -x^2 - 4x - 16
4y + __ = -(x^2 + 4x + __) - 16
[What number do you need to fill in the blank to make a square?]
4y - 4 = -(x^2 + 4x + 4) - 16
[Since you added 4 to the inside of your parentheses, and that quantity is multiplied by the -1 out front, you've essentially added -4 to the right side of the equation. That's why you add a -4 to the left.]
4y = -(x + 2)^2 -12 [Adding 4.]
y = -(1/4)(x + 2)^2 -3 [Dividing by 4.]
Your equation is now in the standard form for parabolas: y = a(x - h)^2 + k
The vertex (h, k) is (-2, -3).
The focus and directrix are 1/(4a) units away from the vertex.
1/(4a) = 1/[4(-1/4)] = 1/-1 = -1.
The focus is 1 unit below the vertex, and the directrix is a unit above it.
Focus: (-2, -4)
directrix: x = -2
2006-06-26 06:21:20 · answer #1 · answered by Anonymous · 0⤊ 0⤋
x^2+4x+4y+16 = 0
x^2+4x+4 + 4y+12 = 0
(x+2)^2 + 4(y+3) = 0
(x+2)^2 = -4(y+3)
Vertex is (-2,-3)
Focus is (-2,-1/16)
Directrix is y=-3-47./16
2006-06-26 13:15:57 · answer #2 · answered by ag_iitkgp 7 · 0⤊ 0⤋
-4y=x²+4x+16
y=-1/4(x+2)² - 3
Talkin' about (-2,-3) for vertex
f=-1/4a; a = -1/4
f=1
so the focus is (-2,-4)
directrix is y= -2
2006-06-26 13:16:33 · answer #3 · answered by bequalming 5 · 0⤊ 0⤋
Leave y on one side.
Use the formulae about the parabola from your book or notebook.
I
2006-06-26 13:40:09 · answer #4 · answered by iyiogrenci 6 · 0⤊ 0⤋