6. The Hyperbola

Cooling towers for a nuclear power plant have a hyperbolic cross-section.
[Image source.]

How do we create a hyperbola?

Take 2 fixed points A and B and let them be 4a units apart. Now, take half of that distance (i.e. 2a units).

Now, move along a curve such that from any point on the curve,

(distance to A) − (distance to B) = 2a units.

The curve that results is called a hyperbola. There are two parts to the curve.

Example 1

Let the distance between our points A and B be 4 cm. For convenience, let's place our fixed points A and B on the number line at (0, 2) and (0, -2), so they are 4 units apart. In this case, a = 1 cm and 2a = 2 cm.

On this page...

The Equation of a Hyperbola

East-West Opening Hyperbola

Definition of a Hyperbola

More Forms of the Equation of a Hyperbola

Even More Forms of the Equation of a Hyperbola

Applications of Hyperbolas

See some interesting Applications [external site]

Now we start tracing out a curve such that P is a point on the curve and the (distance PB) minus (distance PA) is 2 cm. We start at (0, 1).

If we continue, we obtain:

Now, continuing our curve on the left side of the axis gives us:

We also have another part of the hyperbola on the opposite side of the x-axis, this time using:

distance PA − distance PB = 2

We observe that the curves become almost straight near the extremities. In fact, the lines y = x and y = -x (the red dotted lines below) are asymptotes:

[An asymptote is a line that forms a "barrier" to a curve. The curve gets closer and closer to an asymptote, but does not touch it.]

In Example 1, the points (0, 1) and (0, -1) are called the vertices of the hyperbola.

The Equation of a Hyperbola

For the hyperbola with a = 1 that we graphed above in Example 1, the equation is given by:

y² − x² = 1

Notice that it is not a function, since for each x-value, there are two y-values.

We call this example a "north-south" opening hyperbola.

East-West Opening Hyperbola

By reversing the x- and y-variables in our example above, we obtain the following equation.

Example 2

x² − y² = 1

This gives us an "East-West" opening hyperbola, as follows. Our curve passes through -1 and 1 on the x-axis and once again, the asymptotes are the lines y = x and y = -x.

Technical Definition of a Hyperbola

A hyperbola is the locus of points where the difference in the distance to two fixed foci is constant.

This technical definition is one way of describing what we were doing in Example 1, above.

Application of Hyperbolas

Throw 2 stones in a pond. The resulting concentric ripples meet in a hyperbola shape.

More Forms of the Equation of a Hyperbola

There are a few different formulas for a hyperbola.

Considering the hyperbola with centre (0, 0), the equation is either:

1. For a north-south opening hyperbola:

The slopes of the asymptotes are given by:

2. For an east-west opening hyperbola:

The slopes of the asymptotes are given by:

In the examples given above, both a and b were equal to 1, so the slopes of the asymptotes were simply ± 1 and our asymptotes were the lines y = x and y = -x.

What effect does it have if we change a and b?

Example 3

Sketch the hyperbola

Answer

Loading...

Even More Forms of the Equation of a Hyperbola

1. Possibly the simplest equation of a hyperbola is given in the following example.

Example 4 - Equilateral Hyperbola

xy = 1

This is known as the equilateral or rectangular hyperbola.

Notice that this hyperbola is a "north-east, south-west" opening hyperbola. Compared to the other hyperbolas we have seen so far, the axes of the hyperbola have been rotated by 45°. Also, the asymptotes are the x- and y-axes.

---

2. Our hyperbola may not be centred on (0, 0). In this case, we use the following formulas:

For a "north-south" opening hyperbola with centre (h, k), we have:

For an "east-west" opening hyperbola with centre (h, k), we have:

Example 5 - Hyperbola with Axes Shifted

Sketch the hyperbola

Answer

Loading...

---

3. We could expand our equations for the hyperbola into the following form:

Ax² + Bxy + Cy² + Dx + Ey + F = 0 (such that B² > 4AC)

In the earlier examples on this page, there was no xy-term involved. As we saw in Example 4, if we do have an xy-term, it has the effect of rotating the axes. We no longer have "north-south" or "east-west" opening arms - they could open in any direction.

Example 6 - Hyperbola with Shifted and Rotated Axes

The graph of the hyperbola x² + 5xy − 2y² + 3x + 2y + 1 = 0 is as follows:

We see that the axes of the hyperbola have been rotated and have been shifted from (0, 0).

[Further analysis is beyond the scope of this section. ]

Exercise

Sketch the hyperbola

Answer

4. The Parabola

Why study the parabola?

The parabola has many applications in situations where:

On this page...

Definition of a parabola
Formula of a parabola
Arch Bridges
Horizontal Axis
Shifting the Vertex
Applications

• Radiation needs to be concentrated at one point (e.g. radio telescopes, pay TV dishes, solar radiation collectors) or
• Radiation needs to be transmitted from a single point into a wide parallel beam (e.g. headlight reflectors).

Here is an animation showing how parallel radio waves are collected by a parabolic antenna.

http://www.intmath.com/Plane-analytic-geometry/4_Parabola.php

The parabola is defined as the locus of a point which moves so that it is always the same distance from a fixed point (called the focus) and a given line (called the directrix).

In the following graph,

• The focus of the parabola is at (0, p).
• The directrix is the line y = -p.
• The focal distance is |p| (Distance from the origin to the focus, and from the origin to the directrix. We take absolute value because distance is positive.)
• The point (x, y) represents any point on the curve.
• The distance d from any point (x, y) to the focus (0, p) is the same as the distance from (x, y) to the directrix.

[The word locus means the set of points satisfying a given condition.
See some background in Distance from a Point to a Line.]

The Formula for a Parabola - Vertical Axis

Adding to our diagram from above, we see that the distance d = y + p.

Now, using the distance formula on the general points (0, p) and (x, y), and equating it to our value d = y + p, we have

Squaring both sides gives:

(x − 0)² + (y − p)² = (y + p)²

Simplifying gives us the formula for a parabola:

x² = 4py

In more familiar form, with "y = " on the left, we can write this as:

where p is the focal distance of the parabola.

---

Let's play with this in LiveMath. You can change the focal distance to see the effect it has on the parabola's shape.

LIVEMath

---

Now lets animate this, for positive values of p.

LIVEMath

---

Now let's see what "the locus of points equidistant from a point to a line" means in this LiveMath interactive.

LIVEMath

The LiveMath graph is similar to the following.

Each of the colour-coded line segments is the same length in this spider-like graph:

Example - Parabola with Vertical Axis

Need Graph Paper?

Download graph paper

Sketch the parabola

Find the focal length and indicate the focus and the directrix on your graph.

Answer

Loading...

Arch Bridges − Almost Parabolic

The Gladesville Bridge in Sydney, Australia was the longest single span concrete arched bridge in the world when it was constructed in 1964.

The shape of the arch is almost parabolic, as you can see in this image with a superimposed graph of y = −x² (The negative means the legs of the parabola face downwards.)

[Actually, such bridges are normally in the shape of a catenary, but that is beyond the scope of this chapter.]

Parabolas with Horizontal Axis

We can also have the situation where the axis of the parabola is horizontal:

In this case, we have the relation:

y² = 4px

[In a relation, there are two or more values of y for each value of x. On the other hand, a function only has one value of y for each value of x.]

Example - Parabola with Horizontal Axis

Sketch the curve and find the equation of the parabola with focus (-2,0) and directrix x = 2.

Answer

Loading...

Shifting the Vertex of a Parabola from the Origin

This is a similar concept to the case when we shifted the centre of a circle from the origin.

To shift the vertex of a parabola from (0, 0) to (h, k), each x in the equation becomes (x − h) and each y becomes (y − k).

So if the axis of a parabola is vertical, and the vertex is at (h, k), we have

(x − h)² = 4p(y − k)

Let's see what this means in LiveMath:

LIVEMath

If the axis of a parabola is horizontal, and the vertex is at (h, k), the equation becomes

(y − k)² = 4p(x − h)

Exercises

1. Sketch x² = 14y

Answer

Loading...

---

2. Find the equation of the parabola having vertex (0,0), axis along the x-axis and passing through (2,-1).

Answer

Loading...

---

3. We found above that the equation of the parabola with vertex (h, k) and axis parallel to the y-axis is

(x − h)² = 4p(y − k).

Sketch the parabola for which (h, k) is (-1,2) and p = -3.

Answer

Loading...

See also: How to draw y² = x - 2?

Applications of Parabolas

Application 1 - Antennas

A parabolic antenna has a cross-section of width 12 m and depth of 2 m. Where should the receiver be placed for best reception?

Answer

Loading...

Application 2 - Projectiles

A golf ball is dropped and a regular strobe light illustrates its motion as follows...

We observe that it is a parabola. (Well, very close).

What is the equation of the parabola that the golf ball is tracing out?

Answer

3. The Circle

Also on this page:

General form of a Circle

The circle with centre (0, 0) and radius r has the equation:

x² + y² = r²

The circle with centre (h, k) and radius r has the equation:

(x − h)² + (y − k)² = r²

Example 1

Sketch the circle x² + y² = 4.

Answer

Loading...

Example 2

Sketch the circle (x − 2)² + (y − 3)² = 16

Answer

Loading...

Example 3

Sketch the circle (x + 4)² + (y − 5)² = 36

Answer

Loading...

The General Form of the Circle

An equation which can be written in the following form (with constants D, E, F) represents a circle:

x² + y² + Dx + Ey + F = 0

This is called the general form of the circle.

Example:

Find the centre and radius of the circle

x² + y² + 8x + 6y = 0

Sketch the circle.

Answer

Loading...

Exercises

1. Find the equation of the circle with centre (3/2, -2) and radius 5/2.

Answer

Loading...

2. Determine the centre and radius and then sketch the circle:

3x² + 3y² − 12x + 4 = 0

Answer

Loading...

3. Find the points of intersection of the circle

x² + y² − x − 3y = 0

with the line

y = x − 1.

Answer

Perpendicular Distance from a Point to a Line

(BTW - we don't really need to say 'perpendicular' because the distance from a point to a line always means the shortest distance.)

This is a great problem because it uses all these things that we have learned so far:

Later, on this page...

Example using perpendicular distance formula

Need Graph Paper?

• distance formula
• slope of parallel and perpendicular lines
• rectangular coordinates
• different forms of the straight line
• solving simultaneous equations

The distance from a point (m, n) to the line Ax + By + C = 0 is given by:

There are some examples using this formula following the proof.

Proof of the Perpendicular Distance Formula

Let's start with the line Ax + By + C = 0 and label it DE. It has slope .

We have a point P with coordinates (m, n). We wish to find the perpendicular distance from the point P to the line (that is, distance PQ).

We now do a trick to make things easier for ourselves (the algebra is really horrible otherwise). We construct a line parallel to DE through (m, n). This line will also have slope , since it is parallel to DE. We will call this line FG.

Now we construct another line parallel to PQ passing through the origin.

This line will have slope , because it is perpendicular to DE.

Let's call it line RS. We extend it to the origin (0, 0).

We will find the distance RS, which I hope you agree is equal to the distance PQ that we wanted at the start.

Since FG passes through (m, n) and has slope , its equation is or .

Line RS has equation

Line FG intersects with line RS when

Solving this gives us

So after substituting this back into , we find that point R is

Point S is the intersection of the lines and Ax + By + C = 0 (which can be written ).

This occurs when (that is, we are solving them simultaneously)

Solving for x gives

Finding y by substituting back into

gives

So S is the point

So the distance RS, using the distance formula,

is

The absolute value sign is necessary since distance must be a positive value, and certain combinations of A, m , B, n and C can produce a negative number in the numerator.

So the distance from the point (m, n) to the line Ax + By + C = 0 is given by:

Example 1

Find the perpendicular distance from the point (5, 6) to the line -2x + 3y + 4 = 0, using the formula we just found.

Answer

Loading...

Example 2

Find the distance from the point (-3, 7) to the line

Answer