1) 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

1) 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.

The original graph is a circle with a center at  (3, 5)  and a radius of 4 units.

If it is reflected across the line y = 2, the center of the new circle is (3, -1) and the equation is given by

(x - 3)^2 + ( y + 1)^2 = 16      expanding this we have

x^2 - 6x + 9 + y^2 + 2y + 1 = 16     simplify

x^2 - 6x + y^2 + 2y + 10  = 16

x^2 - 6x + y^2 + 2y - 6 = 0    and B = -6  and D = 2  and F = -6     so........B + D + F   =  -6 + 2 - 6 =  -10

Here's a graph of the situation : https://www.desmos.com/calculator/qjdm7ag3vb

2) 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 P is a parabola with focus (4,3) and directrix y=1. The point (8,6) is on P because (8,6) is 5 units away from both the focus and the directrix.
If we write the equation whose graph is P in the form ax^2+bx+c, then what is a+b+c?

The vertex  of this parabola will lie at    (4,2)

And we have this form

y = a(x - h)^2 + k       and  (h,k)  = the vertex = (4,2)   and since (8,6) is on the graph, we can solve for "a" thusly

6 = a(8 - 4)^2 + 2

6 = a(4)^2 + 2

4 = 16a

a = 1/4

So....our equation is

y = (1/4)(x - 4)^2 + 2        expand and simplify

y = (1/4) [x^2 - 8x + 16] + 2

y = (1/4)x^2 -2x + 4 + 2

y = (1/4)x^2 - 2x + 6    and a = (1/4) ,  b = -2   and c = 6  ...so  ...  a + b + c =  (1/4) - 2 + 6 =  4 + 1/4  = 17/4

Here's a graph :  https://www.desmos.com/calculator/htcmqubuwp