MA 108-001 Worksheet Section 11.2 Solutions
Name:
This worksheet is due Friday April 27 in class. Attach your solutions on separate paper.
This is a homework assignment and so it is OK to work with your classmates, but you
must write up your own solutions.
Then there are four types of Parabolas with vertex (0, 0) and with focus on an axis.
Assume a > 0:
Vertex
(0, 0)
(0, 0)
(0, 0)
(0, 0)
Focus
(a, 0)
(−a, 0)
(0, a)
(0, −a)
Directrix
x = −a
x=a
y = −a
y=a
Equation
y 2 = 4ax
y 2 = −4ax
x2 = 4ay
x2 = −4ay
Direction
Opens Right
Opens Left
Opens Upward
Opens Downward
Axis of Symmetry
x-axis
x-axis
y-axis
y-axis
For Problems 1-4, find the equation of the parabola described. These parabolas all have
vertex (0, 0). Find the two points that define the latus rectum, and graph the equation
by hand. In the graph, label the focus, directrix, vertex, and the two points defining
the latus rectum.
Problem 1
Focus at (4, 0).
We know the axis of symmetry goes through the focus and the vertex. Both F and V
are on the x-axis so this parabola is of the form either y 2 = 4ax or y 2 = −4ax. Since
F is to the right of V we know that this parabola opens to the right, and must have
equation of the form y 2 = 4ax. The focus for this type of parabola (see chart) is given
by (a, 0). So a = 4 and the equation is y 2 = 4(4)x = 16x. The directrix is the line
x = −a which is x = −4 in this case. The latus rectum is the line segment through the
focus with endpoints on the parabola. So we plug in the x-coordinate of the focus for
x in the equation of the parabola to see what the y-coordinates of the two endpoints of
the latus rectum are.
y 2 = 16(4) ⇐⇒ y = ±8
So the two endpoints of the L. R. are (4, −8) and (4, 8).
Problem 2
Axis of symmetry the x-axis, containing the point (2, 3).
We did this problem in class Thursday. Here is what the picture should look like.
Problem 3
Focus at (0, −7).
This parabola opens downward because the vertex is on the y-axis, and is below the
vertex (0, 0). So the equation is of the form x2 = −4ay. In this case the focus is (0, −a),
so a = 7 and we have x2 = −4(7)y = −28y. The directrix is y = 7. The latus rectum is
defined by points on the parabola with the same y-coordinate as the focus, and x-coords
as solutions to the equation
x2 = −28(−7) = 196 ⇐⇒ x = ±14
So the endpoints of the L.R. are (−14, −7) and (14, 7).
Problem 4
Axis of symmetry the y-axis, containing the point (3, 4).
Any parabola with axis of symmetry the y-axis opens up or down. There is a point
(3, 4) on the parabola above the vertex (0, 0) so the parabola opens up and has equation
of the form x2 = 4ay. Then to find a we plug in the values x = 3 and y = 4 to the
equation of the parabola.
32 = 4(a)(4) ⇐⇒ 9 = 16a ⇐⇒ a =
9
16
9
9
y = y. The directrix is given by y = −a so
So our parabola has equation
=4
16
4
9
9
in this case D is the line y = − 16
. The focus is the point (0, a) = (0, 16
). The endpoints
of the L. R :
9 9
81
9
2
x =
=
⇐⇒ x = ±
4 16
64
8
x2
Here is the picture:
For Problems 5-8 we find the parabola from the given equation. The vertex of each
parabola is (0, 0). Graph the equation by hand. In the graph, label the focus, directrix,
vertex, and the two points defining the latus rectum.
Problem 5
y 2 = −20x
Comparing y 2 = −20x to our chart, we have a parabola with equation y 2 = −4ax. This
parabola opens to the left, has focus (−a, 0) and directrix given by the line x = a. We
have −20x = −4ax ⇐⇒ −20 = −4a ⇐⇒ a = 5. So the focus is (−5, 0) and the
directrix is the line x = 5. The endpoints of the L. R. have coordinates (−5, 2a) and
(−5, −2a) e.g. (−5, 10) and (−5, −10).
Problem 6
x2 = 6y
3
This equation looks like x2 = 4ay. Since 4ay = 6y we have 4a = 6 so that a = . Focus
2
3
3
then is 0,
, the directrix is given by the line y = − , and the endpoints of the L.R.
2
2
3
3
have coordinates (2a, a) and (−2a, a) which is −3,
and −3,
.
2
2
Problem 7
y2 = x
1
This parabola opens to the right with equation of the form y 2 = 4ax. 4a = 1 so a = .
4
1
1
1 1
1 1
Focus is
, 0 , directrix given by x = − , L.R. endpoints are
,
and
,−
4
4
4 2
4 2
Problem 8
x2 = −y.
This parabola opens
downward
with equation of the form x2 = −4ay.
4a = 1 so
1
1
1
1 1
a = . Focus is 0, − , directrix given by y = , L.R. endpoints are − , −
and
4
4
2 4
4 1 1
,−
2 4