Question

Derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1.

Answer 1

(x-h)^2=4p(y-k)
we know it is this one because the directix is y=something and not x=something

(h,k) is the vertex
p is the distance from the vertex to the focus
it is 1/2 the distance from the vertex to the directix

if p is positive, the focus is above the vertex
if p is negative, the focus is below the vertex


we have
focus is (6,2) and directix is y=1
distance from (6,2) to y=1 is the distance from 2 to 1 which is 1
1/2=p
since 1<2, we see that the focus is above the vertex (when the focus is greater than the directix, then the graph opens to the right or up)

p=1/2
1/2 below (6,2) is (6,1.5)
vertex is (6,1.5)
p=1/2

(x-6)^2=4(0.5)(y-1.5)
(x-6)^2=2(y-1.5)
(x-6)^2=2y-3
2y=(x-6)^2+3
divide both sides by 2
y=1/2 times (x-6)^2+3/2
f(x)=1/2 times (x-6)^2+3/2

2nd option is answer

Answer 2

Answer:

f(x) = 1/2 times (x − 6)^2 + 3/2

Step-by-step explanation:

The definition of a parabola states that any point of the parabola is at the same distance from the focus and from the directrix.  Taking focus at (h, k) and a directrix at y = d:

distance between point (x, y) and the focus = sqrt((x - h)^2 + (y - k)^2)

distance between point (x, y) and the directrix = sqrt((y - d)^2)

Equating them and eliminating the square root:

(x -h)^2 + (y - k)^2 = (y - d)^2

Expanding the y-terms and isolating y

(x -h)^2 + y^2 - 2yk + k^2 = y^2 - 2yd + d^2

(x -h)^2 + k^2 - d^2 = 2y(k - d)

(x -h)^2 + (k - d)(k + d) = 2y(k - d)

(x -h)^2/(2(k - d)) + (k + d)/2 = y

Here h =6, k = 2, d = 1 and y = f(x). Replacing

f(x) = (x -6)^2/(2(2 - 1)) + (2 + 1)/2

f(x) = (x -6)^2/2 + 3/2