Question 601951: If i have a parabola, with the vertex (h,k) at point (-1,-5) and the other 2 points, (not the foci) at (1,-4) and (-3,-4), what would the ending equation be? I do not know how wide it would be, and I can't figure that out. I also have a hyperboa, that I don't even know where to begin with that. HELP PLEASE.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! If i have a parabola, with the vertex (h,k) at point (-1,-5) and the other 2 points, (not the foci) at (1,-4) and (-3,-4), what would the ending equation be? I do not know how wide it would be, and I can't figure that out. I also have a hyperboa, that I don't even know where to begin with that. HELP PLEASE.
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As with all conics, you must know the standard form of equation for the conic and know how to convert it to this form from the given data. Usually, the first thing you must do is complete the square.
For given problem:
Standard form of equation for the parabola: y=A(x-h)^2+k, (h,k)=(x,y) coordinates of vertex, A is a multiplier which affects the slope or steepness of the curve.
The problem here is to find A.
You already have h and k of the formula.
To solve for A, set up an equation by plugging in the coordinates of either one of the given points, (1,-4).
..
y=A(x-h)^2+k
-4=A(1+1)^2-5
-4=4A-5
4A=1
A=1/4
equation:
y=(1/4)(x+1)^2-5
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