for the equation -4x-8y^2=0,give the focus and directrix.then sketch the graph, showing both the focus and the directrix. please show all the step thank you.
-4x-8y^2 = 0
-4x = 8y^2
X = -2y^2
This is an X-parabola; it opens to the left because "a" is negative.
K = -b/2a = 0/-4 = 0
h = 0^2 = 0
V(h,K) = V(0,0).
D(X1,0), V(0,0), F(X2,0).
X1 = h - |1/4a| = 0 - 1/8 = -1/8.
X2 = h + |1/4a| = 0 + 1/8 = 1/8.
Use the following points for graphing:
(X,Y)
(-8,-2)
(-2,-1)
V(0,0)
(-2,1)
(-8,2)
CORRECTION:
F(X1,0), V(0,0), D(X2,0).
To find the focus and directrix of the equation -4x - 8y^2 = 0 and sketch the graph, we can follow these steps:
Step 1: Rewrite the equation in the standard form for a parabola.
The given equation -4x - 8y^2 = 0 can be rearranged to isolate x:
-4x = 8y^2
x = -2y^2
The equation is now represented in the standard form: x = -2y^2.
Step 2: Identify the key components of the parabola and find the vertex.
For the equation x = -2y^2, we notice that the y variable is squared, meaning the parabola opens horizontally. The vertex form of a parabola opening horizontally is (h, k), where h represents the horizontal shift and k represents the vertical shift.
In this case, there is no horizontal shift (h = 0), and the vertex is at (0, 0).
Step 3: Compute the focus.
The formula for the focus of a parabola opening horizontally is given by (h + p, k), where p is the distance from the vertex to the focus. In this case, since h = 0, the formula simplifies to (p, k).
To find p, we can use the formula p = 1 / (4a), where a is the coefficient of y^2 in the standard form equation.
In our equation, the coefficient of y^2 is -2, so a = -1/2.
p = 1 / (4 * (-1/2))
p = 1 / (-2)
p = -1/2
The focus is located at (-1/2, 0).
Step 4: Determine the equation of the directrix.
For a parabola opening horizontally, the equation of the directrix is given by the equation x = h - p, where h is the horizontal shift and p is the distance from the vertex to the focus.
In this case, since h = 0, the formula simplifies to x = -p.
The directrix equation is then x = -(-1/2) = 1/2.
Step 5: Sketch the graph, showing the focus and directrix.
To sketch the graph, use the vertex (0, 0), the focus (-1/2, 0), and the directrix x = 1/2. Remember that the parabola opens horizontally.
The graph will have a shape similar to a sideways "U" with the focus on the left side and the directrix on the right side.
Here is a rough sketch of the graph:
Directrix (x=1/2)
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______V|vertex (0,0)|F (-1/2,0)
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Please note that this is a basic sketch and not to scale. A graphing tool or software can provide a more accurate representation.