Focus and Directrix

Question

Write an equation of the parabola with the given characteristics:

1. opens down; vertex (-1,-3) is 2 units from the focus
2. opens right; vertex (2,4) is 1 unit from the directrix
3. axis of symmetry: x = 0; focus is 8 units above the directrix y = -10
4. focus: (0, 6/7) vertex: (0,0)

Answer 1

conic form for standard parabola

4p(y-k)=(x-h)^2 where p is focal distance and h and k are coordinates of the vertex

1) p=-2

4(-2)(y+3)= (x+1)^2

y+3= -(1/8)(x+1)^2

y= (-1/8)(x+1)^2 -3

2) For a parabola on its side formula is

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

p=1 focal distance is the same as distance from directrix to vertex

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

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

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

3) The distance from the directrix to the vertex equals the distance from the vertex to the focus. If the focus is 8 above the vertex then the distance from the vertex to the directrix must also be 8 which puts the vertex at (0,-2) with p=8 That gives us

4(8)(y+2)= x^2

y= (1/32)x^2 -2

I will leave the last one for you