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 y=ax^2 + bx + c, then what is a+b+c?

Guest Jul 20, 2020

#1**+1 **

I've answered the first one elsewhere today........

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 y=ax^2 + bx + c, then what is a*b*c ?

The vertex will be found at (4,2) and we can write that :

(y - 2) = (a)(x - 4)^2 and since (8,6) is on the curve, we can solve for a

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

4 = (a)(4)^2

4 = (a)16 → a = 4/16 = 1/4

Since the x coordinate of the vertex is given by -b/ (2a) we have that -b/[2 (1/4)] = 4 → -b / (1/2) = 4 → -b = 2 → b = -2

And using the fact that (4,2) is on the graph, we can find c, thusly :

y = ax^2 + bx + c ......so......

2 = (1/4)(4)^2 -2(4) + c

2 = 4 - 8 + c

2 = -4 + c

c = 6

Then a*b*c = (1/4) (-2) (6) = (1/4)(-12) = -3

Guest Jul 20, 2020

#2**+1 **

You are given the directrix and the focus you should be able to deduce that the vertex (between the two)

is at 4,2 this is h,k

Parabola in vetex form y = a(x-h)^2 + k

y = a(x-4)^2 + 2 Sub in th epoint given to calc 'a'

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

a = 1/4

now your equation becomes y = 1/4 (x-4)^2 + 2 expand

= 1/4 x^2 -2x+6 now you can see what a b c are to add them.....

ElectricPavlov Jul 20, 2020