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

y=1/2(x-6)^2+3/2

Step-by-step explanation:

General equation of parabola centered at (h,k) and axis of symmetry is parallel to y-axis is given as

(x-h)^2=4p(y-k)^2

where

vertex is at (h,k)

p shows the distance of focus from vertex, f = (h, k + p)

and directrix is given at y = k - p

if value of p is >0 then the focus is above vertex

if value of p<0 then focus is below vertex

Given:

focus= (6,2)

directrix y=1

Comparing above with standard formula of parabola, we get

Also distance p is half the distance from the (6,2) to the directix y=1 which is  

1/2=p

As p>0, that means given parabola focus lies above vertex

hence parabola opens upwards

Finding vertex of parabola:

As focus is is at (6,2) and also 1/2 above the vertex so we get vertex at

V=(6,2-1/2)

V=(6,1.5)

Now by comparing above with the standard formula of parabola we have

h=6, k=1.5

Putting values in the formula we get

(x-6)^2=4(1/2)(y-1.5)

(x-6)^2=2(y-1.5)

(x-6)^2=2y-3

Or by re-arranging the terms equation can also be written as

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

answer: ask

step-by-step explanation: