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Mathematics 10 Online
OpenStudy (anonymous):

what is the equation of a parabola with focus at F (2,-2) and directrix at y=4

OpenStudy (anonymous):

@jim_thompson5910 Please help!

OpenStudy (anonymous):

@dan815 help please!

OpenStudy (anonymous):

I need to find the equation

jimthompson5910 (jim_thompson5910):

The y coordinate of the focus is y = -2 how far is y = -2 from y = 4 ?

OpenStudy (anonymous):

6?

jimthompson5910 (jim_thompson5910):

correct

OpenStudy (anonymous):

y = 1/12 x^2 -1/3x-2/3?

jimthompson5910 (jim_thompson5910):

cut this in half to get 6/2 = 3 the distance from the focus to the vertex is 3 units, so p = 3 (p is the focal distance)

jimthompson5910 (jim_thompson5910):

what is the midpoint of (2,-2) and (2,4) ?

OpenStudy (anonymous):

4?

jimthompson5910 (jim_thompson5910):

plot the two points (2,-2) and (2,4) which point is directly in the middle of the two of them?

OpenStudy (anonymous):

2,1?

jimthompson5910 (jim_thompson5910):

that's the vertex (h,k) = (2,1)

jimthompson5910 (jim_thompson5910):

we will plug p = 3, h = 2, k = 1 into the "vertical parabola" formula shown here http://www.mathwords.com/f/focus_parabola.htm

OpenStudy (anonymous):

y = -1/12 x^2 -1/3x-2/3?

jimthompson5910 (jim_thompson5910):

4p(y-k) = (x-h)^2 4*3(y-1) = (x-2)^2 12(y-1) = (x-2)^2 y-1 = (1/12)*(x-2)^2 y = (1/12)*(x-2)^2+1

jimthompson5910 (jim_thompson5910):

one final adjustment the parabola opens downward, so the leading coefficient 1/12 needs to be negative so the actual equation is y = (-1/12)*(x-2)^2+1

jimthompson5910 (jim_thompson5910):

hopefully you can see how I isolated y

OpenStudy (anonymous):

thanks so it is y = -1/12 x^2 +1/3x+2/3?

jimthompson5910 (jim_thompson5910):

yes it is once you expand out y = (-1/12)*(x-2)^2+1

OpenStudy (anonymous):

thanks can you help with a few more?

jimthompson5910 (jim_thompson5910):

I'll do 2 more

OpenStudy (anonymous):

a parabola has a focus at F(-2,4)and a directrix at y=0. what is the location of the vertex?

OpenStudy (anonymous):

what formula should be used to derive the equation of a parabola given its focus and directrix

OpenStudy (anonymous):

what is the general equation for a parabola with its vertex at the origin?

jimthompson5910 (jim_thompson5910):

we will use 4p(y-k) = (x-h)^2 again p = focal distance (h,k) = vertex

jimthompson5910 (jim_thompson5910):

to find the focal distance, you find the distance from directrix to focus, then you cut that distance in half

jimthompson5910 (jim_thompson5910):

so what is the focal distance?

OpenStudy (anonymous):

-2?

jimthompson5910 (jim_thompson5910):

y = 0 is the directrix y = 4 is the y coordinate of the focus what is the change in y?

OpenStudy (anonymous):

2?

OpenStudy (anonymous):

(-1,4)

jimthompson5910 (jim_thompson5910):

y = 0 to y = 4 is a change of 4

jimthompson5910 (jim_thompson5910):

4-0 = 4

jimthompson5910 (jim_thompson5910):

cut this distance in half 4/2 = 2 so p = 2

jimthompson5910 (jim_thompson5910):

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jimthompson5910 (jim_thompson5910):

|dw:1431650099162:dw|

jimthompson5910 (jim_thompson5910):

|dw:1431650182362:dw|

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