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 \(B+D+F\).
Please somebody!
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 \(a+b+c\).
+310 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 \(x^2-6x+y^2+2y-6=0\).
\(B=-6, D=2, F=-6\)
Add these up to get your answer.
