Find the standard form of the equation of the parabola with the given characteristics

Question

Find the standard form of the equation of the parabola with the given characteristics.

a) Vertex: (2, −8); focus: (5, −8)

b) Vertex: (1, 4); directrix: y = −5

Answer 1

a) focus is to the right of the vertex, which means the parabola opens to the right with general formula

x-h = a(y-k)^2 where (h,k) is the vertex

x-2 = a(y+8)^2

to calculate a find another point on the parabola, directly, vertically above the focus

the directrix is the same distance to the left of the vertex as the focus is to the right

directrix = -1

the point on the parabola vertically above the focus has the same x coordinate as the focus:5

distance to the directrix = 6 (5--1) = vertical distance above the focus -8+6 = -2

the point on the parabola is (5,-2) plug it into the general formula and solve for 1

x-2 = a(y+8)^2

5-2 = a(-2+8)^2

3 = 36a

a = 3/36 = 1/12

the parabola equation is

x-2 = 1/12)(y+8)^2

b) the directrix is below the vertex, so this parabola opens upwards. Focus is (1,13) equidistant with the direction to the vertex

y-h= a(x-k)^2 with (k,h) = (1,4)

y-4 = a(x-1)^2

find a point on the parabola horizontally from the focus, 9 units away (10,13)

plug that point in to solve for 1

13-4 =81a

a = 9/81 = 1/9

y-4 = (1/9)(x-1)^2

Answer 2

a) If you plot those two points you see the focus is to the right of the vertex, so the parabola would open horizontally and the equation is of the form (y-k)^2 = 4p(x-h), where (h,k) is the vertex (2,8) and p = -5. (y-8)^2 = -20(x-2).

b) directrix y = -5, so parabola would open vertically, p = 5, and the equation is of the form (x-h)^2 = 4p(y-k), where (h,k) = (1,4). (x-1)^2 = 20(y-4).