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Question 1166292: In 1 and 2, do the following:
a) Write the equation of the parabola in standard form.
b) Find the vertex, focus, equation of the directrix and axis of symmetry of the parabola.
1. x^2-6x-5y-1=0
2. y^2-2y+3x-8=0
In 3 and 4, find an equation of the parabola that satisfies the given conditions. Express the answer in standard and general forms.
3. Vertex at V(-4,3),directrix at y=5.
4. Focus at F(-1,-5),directrix at x=-2.
5. Each cable of a suspension bridge is in the shape of a parabola and is supported by two towers at each end (refer to the figure below). The shape of the cable is modeled by the equation
x^2=200y
where x and y are measured in meters. Find the coordinates of the focus.
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
One problem per post!!!
Here is all you need to work these problems yourself....
Vertex form of equation of a parabola that opens up or down:
or
Vertex form of equation of a parabola that opens right or left:
or
In both formulas, the vertex is (h,k), and p is the directed distance from the directrix to the vertex and from the vertex to the focus.
Here is a problem like your problems 1 and 2....
Complete the square in x and put the equation in vertex form.
That is vertex form: h=2; k=4; 4p=4 so p=1.
The vertex is (h,k) = (2,4).
p=1, so the directed distance from the directrix to the vertex is 1, and the directed distance from the vertex to the focus is 1. That makes the directrix y=3 and the focus (2,5).
Your problem 5 is solved in a similar fashion.
For problems 3 and 4, use the definitions of (h,k) and p to write the equations.
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