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The graph of $$(x-3)^2 + (y-5)^2=16$$ is reflected over the line $$y=2$$. The new graph is the graph of the equation $$x^2 + Bx + y^2 + Dy + F = 0$$ for some constants $$B$$$$D$$, and $$F$$. Find $$B+D+F$$.

Jun 30, 2020

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Also I have this:

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  because  $$\mathcal{P}$$ 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 $$a+b+c$$.

Jun 30, 2020
#2
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This graph is a circle with center (3,5) and radius 4.

The radius of the circle is still going to be the same after reflection, but the center changes to (3,-1).

So our equation is $$(x-3)^2+(y+1)^2=16$$.

Expanding this, we get $$x^2-6x+y^2+2y-6=0$$.

$$B=-6, D=2, F=-6$$