What is an equation of a parabola with vertex at the origin and directrix y =2.8?

Work Shown:

Vertex = (0,0) = (h,k)

directrix is y = 2.8

vertical distance from y = 0 to y = 2.8 is 2.8 units

2.8 = 28/10 = 14/5

p = 14/5 is the focal distance

a = -1/(4*p) = -1/(4*14/5) = -1/(56/5) = -5/56

The value of 'a' is negative because the directrix is above the focus, which means the focal point is below the vertex.

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

y = (-5/56)(x-0)^2 + 0

y = (-5/56)x^2

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

a

step-by-step explanation:

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