What is an equation of a parabola with vertex at the origin and focus (0, 7)

a. y = -1/28 x^2

b. y = 1/28 x^2

c. x = -1/28 y^2

d. x = 1/28 y^2

Explanation:

p = focal distance = 7

the focal distance is from the vertex to the focus

The parabola opens upward (since the vertex is below the focus) so 'a' is positive and a = 1/(4p) = 1/(4*7) = 1/28

The vertex is (h,k) = (0,0)

Put this all together into the formula below and simplify

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

y = (1/28)(x - 0)^2 + 0

y = (1/28)x^2

for finding the midpoint of any given 2 pair of points simply add the respective x coordinates with other x coordinate and the y coordinate with other y coordinate and divide them by 2

in your question it will be :

(3+1)/2 for the x coordinate of mid point i.e 2

[8+(-4)]/2 for the y coordinate of midpoint i.e 2

so coordinates of mid point of points (3,8) and (1,-4) will be (2,2)

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this