|
Question 691514: Find the vertex, focus and directrix and sketch the graph of the parabola from the equation (x+1)^2 + 8(y+2)=0.
First, I got the vertex (-1,-2), h=-1 k=-2. Next I tried to find teh focus but wanted to get "p" first. So 4p=8 p=2
Using the formula to get the focus (h,k+p) I get (-1,-2+2)-(1,0) but when I looked at my answer book it said that p=-2 and the focus is (-1,-4). What may I be doing wrong.
Answer by Edwin McCravy(20064)   (Show Source):
You can put this solution on YOUR website!
The correct standard form of theequation of a parabola with
a vertical axis of symmetry is
(x - h)² = 4p(y - k)
Your parabola equation,
(x + 1)² + 8(y + 2) = 0
is not quite in that form. So we subtract 8(y + 2)
from both sides:
(x + 1)² = -8(y + 2)
Compare that to
(x - h)² = 4p(y - k)
And we see that h = -1, k = -2 and 4p = -8
p = -2
We plot the vertex (h,k) = (-1,-2), and since p = -2, a
negative number, the focus is 2 units BELOW the vertex.
So the focus is (-1,-4). And the directrix is 2
units ABOVE the vertex and therefore has equation y = 0,
which just happens to be the x-axis. The focal chord or
latin rectum is a line through the focus, bisected by the
focus ad which is |4p| = 8 units long. So we draw in the latus
rectum, in green:
and sketch in the parabola:
Edwin
|
|
|
| |
