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Geometrically speaking, a parabola is defined as the set of points that are the same distance from a given point and a given line. The point is called the focus of the parabola and the line is called the directrix of the parabola.

Suppose $$\mathcal{P}$$ is a parabola with focus $$(4,3)$$ and directrix $$y=1$$. The point $$(8,6)$$ is on $$\mathcal{P}$$ because $$(8,6)$$ is 5 units away from both the focus and the directrix.

If we write the equation whose graph is $$\mathcal{P}$$ in the form $$y=ax^2+bx+c$$, then what is $$a+b+c$$?

Apr 13, 2020

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A graphing form for a parabola is:  y - k  =  a(x - h)2 where (h, k) is the vertex.

The vertex is midway between the focus and the directrix.

Since the focus is (4, 3) and the directrix is  y = 1, the vertex is at (4, 2).

Therefore, the graphing form is:  y - 2  =  a(x - 4)2.

To find a, let's put the value of (8, 6) into this formula:  6 - 2  =  a(8 - 4)2

4  =  a(4)2

4  =  a·16

a  =  4/16   --->   a = ¼

So, the equation is:  y - 2  =  ¼(x - 4)2

Multiplying out:         y - 2  =  ¼(x2 - 8x + 16)

y - 2  =  ¼x2 - 2x + 4

y  =  ¼x2 - 2x + 6

a = ¼   b = -2  c = 6

Apr 13, 2020