Math 113 Lecture #37
§10.5: Conic Sections
Parabolas. Fix a point F in the xy-plane and call it the focus.
Fix a line in the xy-plane that does not contain the focus F , and call this line the
directrix.
A parabola is the set of points in the xy-plane that are equidistant (i.e., same distant)
from the focus F and the directrix.
Such a parabola is illustrated in the following graph the focus is the point F = (0, 1),
and the directrix is the line y = −1.
2.0
1.5
1.0
0.5
0.0
!2
!1
x
0
1
2
!0.5
y !1.0
!1.5
!2.0
The halfway point between the focus and the directrix lies on the parabola and is called
the vertex of the parabola.
We obtain a relatively simple Cartesian equation for a parabola when we place the focus
on the y axis, say at the point (0, p) for p > 0, and the directrix parallel to the x-axis,
say y = −p.
For any point P = (x, y) on the parabola, the distance from the P to the focus is
p
|P F | = x2 + (y − p)2 ,
and the distance from the point P to the directrix is
|y + p|.
This two distances are the same for P on the parabola, and so
p
x2 + (y − p)2 = |y + p|.
Now squaring both sides gives
x2 + (y − p)2 = (y + p)2 .
Expanding the squares gives
x2 + y 2 − 2py + p2 = y 2 + 2py + p2 .
Simplifying this gives
x2 = 4py.
This is a Cartesian equation for the parabola with focus (0, p) and directrix y = −p.
This always works if p < 0 too, in which case the parabola opens downwards.
All of this can be done when the focus is the point (p, 0) for p 6= 0 and the directrix is
the line x = −p, which gives
y 2 = 4px
as a Cartesian equation for the parabola, opening to the left when p < 0 and to the right
when p > 0.
Example 1. Find the focus and the directrix for the parabola y + 3x2 = 0.
We put his equation into its x2 = 4py form to identify the value of p which gives the
focus (0, p) and the directrix y = −p:
y
x2 = .
3
This requires that 4p = 1/3, and so p = 1/12.
Ellipses. Fix two distinct points F1 and F2 in the xy-plane, and called the foci.
An ellipse in the xy-plane is the set of points the sum of whose distances from the two
foci is a constant.
Here is an illustration of this for the foci located at the points (−3, 0) and (3, 0).
6
5
4
3
2
1
0
!6
!5
!4
!3
x
!2
!1
0
!1
!2
y !3
!4
!5
!6
1
2
3
4
5
6
To find a Cartesian equation for an ellipse, we place the two foci on the x-axis spaced
equally from the origin, say F1 = (−c, 0) and F2 = (c, 0) for a constant c.
Let P = (x, y) be a point on the ellipse, and let 2a be the distance from P to F1 plus the
distance from P to F2 .
Then
|P F1 | + |P F2 | = 2a.
We use the distance formula to write this in terms of x and y:
p
p
(x + c)2 + y 2 + (x − c)2 + y 2 = 2a.
We eliminate the radicals, which takes two squarings, to get
(a2 − c2 )x2 + a2 y 2 = a2 (a2 − c2 ).
From the triangle with vertices P , F1 , and F2 , we see that 2c is smaller than 2a, and so
c < a.
This implies that a2 − c2 > 0, and we introduce a quantity b by a2 − c2 = b2 .
The equation for the ellipse then becomes
b 2 x 2 + a2 y 2 = a2 b 2 .
Dividing this through by a2 b2 gives
x2 y 2
+ 2 =1
a2
b
as a Cartesian equation for an ellipse with foci at (±c, 0), where c2 = a2 −b2 and a ≥ b > 0.
The x-intercepts of this ellipse are located at (±a, 0) (i.e., set y = 0 and solve for x), and
are called the vertices of the ellipse.
The line segment between this vertices is the longest line between opposite points on the
ellipse and is called the major axis of the ellipse.
Notice that if a = b, then the two foci coalesce into one focus and the ellipse is a circle
of radius a.
If instead we put the foci on the y-axis at (0, ±c), then a Cartesian equation for the
ellipse is
x2 y 2
+ 2 =1
b2
a
2
2
2
where c = a − b and a ≥ b > 0.
The vertices for this ellipse are located at (0, ±a) and the line segment between these
vertices is the major axis for this ellipse.
Example 2. Find the foci and vertices of the ellipse
x2 y 2
+
= 1.
9
5
Since 9 ≥ 5 > 0, the foci are of the form (±c, 0) where c2 = a2 − b2 = 9 − 5 = 4, i.e.,
c = ±2.
The vertices are located on the x-axis at the points (±3, 0).
Hyperbolas. Fix two distinct points in the xy-plane and call them foci.
A hyperbola is the set of points in the xy-plane the difference of whose distances from
the foci is a constant.
This is illustrated for the foci located at (±3, 0).
5
4
3
2
1
0
!5
!4
!3
!2
!1
0
1
2
3
4
5
!1
x
!2
y
!3
!4
!5
We can find a Cartesian equation for a hyperbola with foci at (±c, 0) rather easily.
For a point P = (x, y) on the hyperbola, if we set |P F1 | − |P F2 | = ±2a, then simple
calculations give
x2 y 2
− 2 =1
a2
b
2
2
2
where c = a + b .
Setting y = 0 shows that the points (±a, 0) lie on the hyperbola, and these are called
the vertices of the parabola.
The hyperbola consists of two branches, which can be seen by rewriting the Cartesian
equation for the hyperbola in the form
y2
x2
=
1
+
.
a2
b2
This shows that |x| ≥ |a| on the hyperbola, so one branch is given by x > |a|, and the
other branch by x < −|a|.
If instead we write the Cartesian equation for a hyperbola in the form
y2
x2
=
−1
b2
a2
we see that for large values of x we can ignore the −1 term so that the hyperbola is
approximated by
y2
x2
b
= 2 or y = ± x.
2
b
a
a
The graphs of these lines are called the asymptotes of the hyperbola, as illustrated here.
5
4
3
2
1
0
!5
!4
!3
!2
!1
0
1
2
3
4
5
!1
x
!2
y
!3
!4
!5
If we had put the foci on the y-axis, say at (0, ±c), then we would find that a Cartesian
equation for the hyperbola is
y 2 x2
− 2 =1
a2
b
2
2
2
where c = a +b , for which the vertices are (0, ±a), and the asymptotes are y = ±(a/b)x.
Example 3. Find the vertices, foci, and asymptotes for the hyperbola
y2
x2
−
= 1.
16 36
Setting x = 0 gives the vertices, which are (0, ±4).
√
Since c2 = a2 + b2 = 16 + 36 = 52, the foci are located at (0, ± 52).
The asymptotes are the lines y = ±(a/b)x = ±(16/36)x = ±(4/9)x.
Shifted Conics. All of the Cartesian equations for the conic sections have been centered
about the origin.
To put the center of the conic section somewhere other than the origin, we shift the
Cartesian equation appropriately.
This is done by replacing x with x − h and y with y − k, which puts the center of the
conic section at the point (h, k).