Derive the equation of the parabola with a focus at (6, 2) and a directrix of y = 1.a f(x) = −1/2 (x − 6)2 + 3/2b f(x) = 1/2 (x − 6)2 + 3/2c f(x) = −1/2 (x + 3/2)2 + 6d f(x) = 1/2 (x + 3/2)2 + 6 make sure to provide how you got the answer, i want to know how to do these for future reference. also, make sure to incorporate the distance formula methodd=

B. f(x) = (1/2)(x -6)^2 +3/2

Let's do this in general terms for a focus at (h, k) and a directrix at y = d.

We start with a generic point (x, y) and we want to have it satisfy the relation that its distance from the focus is the same as its distance from the directrix (the definition of a parabola).

... distance from (x, y) to the focus = √((x -h)^2 +(y -k)^2)

... distance from (x, y) to the directrix = √((y - d)^2)

Equating these distances and squaring both sides of the equation, we get ...

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

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

Expanding the y term, we find we can cancel the y^2 terms to get an equation you're more familiar with.

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

... (x -h)^2 +k^2 -d^2 = 2y(k -d) . . . . . . note the difference of squares on the left

Dividing by the coefficient of y gives ...

... y = 1/(2(k-d))·(x -h)^2 +(k+d)/2 . . . . . . equation of a general parabola (opens vertically)

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Sometimes, you will see the difference (k-d) designated as 2p, so the equation is written

... y = 1/(4p)·(x -h)^2 + (k-p) . . . . . p is the distance from focus (h, k) to vertex

And sometimes, (h, k) is defined as the vertex, not the focus, so the equation is written

... y = 1/(4p)·(x -h)^2 +k . . . . . . . (h, k) is the vertex

For focus (6, 2) and directrix y=1, we have (h, k) = (6, 2) and d=1. Using the first bold equation above, our equation for the parabola is ...

... y = 1/(2(2-1))·(x -6)^2 +(2+1)/2

... y = (1/2)(x -6)^2 +3/2 . . . . matches selection B

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50