Solutions
Problem
Set #5
Geometry
1. In each case below, construct the equation of the parabola having the specified characteristics. Also indicate where the
vertex is
a) Focus
at (0,5) and directrix the
line y= -5 solution:
(d=5 here) vertex
at (0,0)
b) Focus at (0,2) and directrix the line
y = -2 solution: 
vertex at (0,0)
c) Focus
at (0,4) and directrix y=0
(the x axis) solution:
vertex at (0,2)
d) Focus
at ( 3,4) and directrix
y = -4 solution: 
d – 4 ,
vertex at (3,0)
e) Focus
at (3,9) and directrix the line y = +1 solution:
d=4, vertex at
(3,5)
f) Vertex
at (2,10) and directrix
3 units below it solution:
d = 3
g) Vertex at
(-3,5) and directrix 7 units above
it Solution: 
note minus sign
as this opens down
2. Based on each equation below, decide
where the vertex is
where the focus is
where the directrix is
whether the parabola opens up or down
a)
20y = x2 vertex at (0,0) d =
5 so focus at (0,5) and y=
-5 is the directrix. Opens up.
b) y = -x2/12 vertex at (0,0)
d = 3 Opens down. focus at (0,-3) and directrix at y = +3
c) y + 3 = (x – 9)2/4 vertex at
(9,-3) opens up. d=1
focus at (9, -2) directrix y= -4
d) y -4 = -(x + 1)2/36 vertex (-1,4)
opens down d=9 focus at (-1,-5) directrix y = 13
3. Use algebraic methods on each equation to put it into “standard form”. Once there, identify where the vertex is and whether the parabola opens up or down.
a.
x2
-8x - 16y = 0 y +16= (1/16)(x -4)2 so vertex at (4, -16) d = 4
focus at (4, -12) opens up
b.
x2 -32y +
64 = 0 y -2 = (1/32)x2 so d = 8
vertex at (0,2) focus at
(0,10) opens up
c.
x2 +
16x - y + 6 = 0
y +58 =
(x+8)2 so
d = 1/4 vertex at (-8,-58)
focus at (-8,-57 3/4) opens up
d.
x2 -10x
+ 32y = 7 y -1 = (-1/32)(x-5)2 so d = 8
vertex at (5,1) focus at
(5,-7) opens down