CONIC SECTIONS (c)
THE ELLIPSE - COMPLEX FORM
An ellipse is the locus of a point that moves so that the sum of its distances to two fixed points is a constant.
If z1 and z2 are two fixed points in the complex plane and a the ends of a length of string are attached at these two
points, then a pencil can be used with the string to trace out an ellipse.
pencil
z1
z2
The length of the string, s, is the same as the length of the major axis and the equation of the ellipse is
|𝒛 − 𝒛𝟏 | + |𝒛 − 𝒛𝟐 | = 𝒔
𝒔 > |𝒛𝟏 − 𝒛𝟐 |
and
Examples
1.
Describe the locus of z (x, y) given that
∴
|𝑧 + 2| + |𝑧 − 2𝑖|
|𝑧 − (−2 + 0𝑖)| + |𝑧 − (0 + 2𝑖)|
|𝑧 + 2| + |𝑧 − 2𝑖| = 5.
=
=
5
5
The locus of z (x, y) is an ellipse with foci at (-2, 0) and (0, 2). The major axis is of length 5 units.
2.
Sketch the locus of a point that moves so that
∴
|𝑧 − 2 + 3𝑖| + |𝑧 + 1 + 𝑖|
|𝑧 − (2 − 3𝑖)| + |𝑧 − (−1 − 𝑖)|
|𝑧 − 2 + 3𝑖| + |𝑧 + 1 + 𝑖| = 6.
=
=
6
6
The locus is an ellipse with foci at (2, -3) and (-1, -1). The centre (0.5, -2) is the midpoint of the two
foci.
(-1, -1)
(0.5, -2)
(2, -3)
3.
An ellipse has real intercepts at ± 8 and imaginary intercepts at ± 3i. What equation describes this
ellipse?
iy
Major axis, s
=
=
2a
16
3
8
The foci and imaginary intercepts form right
triangles at the Origin and z1 and z2 can be found
using Pythagoras’ Theorem:
2
8
z12
z1
=
=
=
2
z2
z1
8
x
-3
2
3 + z1
55
and
√55
The equation of the ellipse is
-8
z2
=
−√55
|𝑧 − √55| + |𝑧 + √55| = 16
If the ellipse is vertical, the same process is used but the foci are on the y-axis instead and an adjustment is made
to the calculation as a result. The major axis is vertical and the right triangle is made between the Origin, the real
intercepts and the foci.
THE HYPERBOLA - COMPLEX FORM
An hyperbola is the locus of a point that moves so that the difference
between its distances from two fixed points is constant. The complex
equation is given by:
|𝒛 − 𝒛𝟏 | − |𝒛 − 𝒛𝟐 | = ±𝒔
and
-
Transverse axis
𝒔 < |𝒛𝟏 − 𝒛𝟐 |
The ±s gives the two branches of the hyperbola. The value of s gives
the shortest distance between the two branches of the hyperbola. It is
the distance along the line through the foci and is known as the
transverse axis. The vertices of the hyperbola are at the ends of the
transverse axis.
Also
iy
z2
z1
x
5
x
Vertices
if s < 0, the branch is closer to z1;
if s > 0, the branch is closer to z2.
Examples
4.
iy
Sketch the hyperbola |𝑧 − 5| − |𝑧 + 5| = ±6 marking all
relevant points.
The equation represents both branches of an hyperbola with
foci at (5, 0) and (-5, 0). The centre will be the midpoint of the
foci – i.e. the Origin.
The transverse axis is 6 units long so the vertices are at (-3, 0)
and (3, 0).
-5 -3
3
iy
5.
Sketch the locus of |𝑧 − 2𝑖| − |𝑧 + 4𝑖| = 5.
2
The equation shows one branch of an hyperbola with foci
at z = 2i and z = - 4i. The centre will be at z = -i.
-1
x
Because s > 0, the one branch will be closer to -4i.
-3.5
-4
The transverse axis is 5 units long, so the one vertex will
be 2.5 units from the centre – i.e. at z = -3.5i.
6.
Sketch the region shown by |𝑧| − |𝑧 + 1| > −12.
This hyperbola has foci at z = 0 and z = -1, the centre is at z = -0.5 and there is only the one branch
which will be closer to z = 0. The vertex will be at z = -0.25 as the transverse axis is 0.5 units long.
Testing a point to see if it is within the required region:
|2| − |2 + 1| > −12
if z = 2 then
so z = 2 is not in the region.
2–3>−
becomes
1
2
which is false,
iy
We will need to shade the region to the left of the curve
and the curve will not be included.
-0.25
-1
7.
0
x
An ellipse and hyperbola share foci at (-3, 0) and (3, 0) as shown in the diagram. Use complex
inequalities to describe the shaded region.
iy
For the hyperbola, the transverse axis is 4 units and
foci are given so it has equation |𝑧 − 3| − |𝑧 + 3| = 4
Testing the point z = 0 gives 3 – 3 = 0, so
-6
|𝑧 − 3| − |𝑧 + 3| ≤ 4
-3 -2
For the ellipse, there is one vertex at z = -6 so the
major axis is 12.
Testing the point z = 0 gives 3 + 3 = 6, so the equation
of the ellipse will be
|𝑧 − 3| + |𝑧 + 3| < 12
Because this is an overlapping region, the shaded area is described by
|𝑧 − 3| − |𝑧 + 3| ≤ 4
∩
|𝑧 − 3| + |𝑧 + 3| < 12
2
3
x
8.
An hyperbola has a focus at z = 2, centre at the Origin and it passes through the point z = 3 - 2√6𝑖. Find
the equation of the curve and hence the length of the transverse axis.
iy
|𝑧 − 2| − |𝑧 + 2| = ±𝑠
The equation will be
where s = length of transverse axis.
If z = 3 - 2√6𝑖, then it is closer to the focus at z = 2,
so s < 0 (i.e. negative):
∴
∴
|3 − 2√6𝑖 − 2| − |3 − 2√6𝑖 + 2| = −𝑠
|1 − 2√6𝑖| − |5 − 2√6𝑖| = −𝑠
|5 − 2√6𝑖| − |1 − 2√6𝑖| = 𝑠
Now,
|5 − 2√6𝑖|
and
So,
|1 − 2√6𝑖|
s
=
√𝑥 2 + 𝑦 2
=
√52 + (2√6)2
=
7
=
√12 + (2√6)2
=
5
=
=
7–5
2
-2
2
x
z = 3 - 2√6𝑖
The equation of the hyperbola is |𝑧 − 2| − |𝑧 + 2| = ±2 and the transverse axis is 2 units long.