Parabolas; find Vertex, focus, directrix? 1) (y+1)^2 = -4(x-2)
the parabola with directrix at x=p and vertex at (0,0), and focus at x = -p is
y^2 = 4px.
You have a parabola with
4p = -4, so p = -1
and it is shifted by (2,-1).
So,
vertex = (2,-1)
focus = (1,-1)
directrix: x = 3
See here for confirmation:
http://www.wolframalpha.com/input/?i=parabola+%28y%2B1%29^2+%3D+-4%28x-2%29
To find the vertex, focus, and directrix of the given parabola (y+1)^2 = -4(x-2), we can use the vertex form of a parabola equation - (y-k)^2 = 4a(x-h) - where (h, k) represents the vertex and "a" determines the shape and orientation of the parabola.
Comparing the given equation to the vertex form, we have h = 2 and k = -1. Therefore, the vertex of the parabola is V(2, -1).
Now, let's find the value of "a." From the equation, we see that -4a = -4. Solving for "a," we get a = 1.
Next, to find the focus, we use the formula F(h+a, k), where (h, k) is the vertex coordinates we found earlier and "a" is the value we just calculated. Plugging in the values, we get F(2+1, -1), which simplifies to F(3, -1). Thus, the focus of the parabola is F(3, -1).
Lastly, we need to determine the directrix. For a parabola in vertex form, the directrix is a horizontal line with the equation y = k - a. Plugging in the values, we get y = -1 - 1, which simplifies to y = -2. Hence, the directrix of the parabola is the line y = -2.
In summary:
- The vertex of the parabola is V(2, -1).
- The focus of the parabola is F(3, -1).
- The directrix of the parabola is the line y = -2.