The ellipse, parabola and hyperbola

1 The ellipse, parabola and hyperbola

Mathematicians, engineers and scientists encounter numerous functions in their work: polynomials, trigonometric and hyperbolic functions amongst them. However, throughout the history of science one group of functions, the conics, arise time and time again not only in the development of mathematical theory but also in practical applications. The conics were first studied by the Greek mathematician Apollonius more than 200 years BC.

Essentially, the conics form that class of curves which are obtained when a double cone is intersected by a plane. There are three main types: the ellipse , the parabola and the hyperbola . From the ellipse we obtain the circle as a special case, and from the hyperbola we obtain the rectangular hyperbola as a special case. These curves are illustrated in the Figures 1 and 2.

Figure 1a :

{Circle and ellipse}

Figure 1b:

{Parabola}

Figure 2 :

{Hyperbola}

1.1 The ellipse

We are all aware that the paths followed by the planets around the sun are elliptical. However, more generally the ellipse occurs in many areas of engineering. The standard form of an ellipse is shown in Figure 3.

Figure 3

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If $a > b$ (as in Figure 1) then the $x$ -axis is called the major-axis and the $y$ -axis is called the minor-axis . On the other hand if $b > a$ then the $y$ -axis is called the major-axis and the $x$ -axis is then the minor-axis. Two points, inside the ellipse are of importance; these are the foci . If $a > b$ these are located at coordinate positions $\pm a e$ (or at $\pm b e$ if $b > a$ ) on the major-axis, with $e$ , called the eccentricity , given by

$e^{2} = 1 - \frac{b^{2}}{a^{2}} (b < a) or by e^{2} = 1 - \frac{a^{2}}{b^{2}} (a < b)$

The foci of an ellipse have the property that if light rays are emitted from one focus then on reflection at the elliptic curve they pass through at the other focus.

Key Point 1

The standard Cartesian equation of the ellipse with its centre at the origin is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

This ellipse has intercepts on the $x$ -axis at $x = \pm a$ and on the $y$ -axis at $\pm b$ . The curve is also symmetrical about both axes. The curve reduces to a circle in the special case in which $a = b$ .

Example 1

Sketch the ellipse $\frac{x^{2}}{4} + \frac{y^{2}}{9} = 1$
Find the eccentricity $e$
Locate the positions of the foci.

Solution

We can calculate the values of

y

x

changes from 0 to 2:

$x$	0	0.30	0.60	0.90	1.20	1.50	1.80	2
$y$	3	2.97	2.86	2.68	2.40	1.98	1.31	0

From this table of values, and using the symmetry of the curve, a sketch can be drawn (see Figure 4). Here $b = 3$ and $a = 2$ so the $y$ -axis is the major axis and the $x$ -axis is the minor axis.

Here $b = 3$ and $a = 2$ so the $y$ -axis is the major axis and the $x$ -axis is the minor axis.

$e^{2} = 1 - a^{2} ∕ b^{2} = 1 - 4 ∕ 9 = 5 ∕ 9 ∴ e = \sqrt{5} ∕ 3$
Since $b > a$ and $b e = \sqrt{5}$ , the foci are located at $\pm \sqrt{5}$ on the $y$ -axis.
Figure 4

Key Point 1 gives the equation of the ellipse with its centre at the origin. If the centre of the ellipse has coordinates $(α, β)$ and still has its axes parallel to the $x$ - and $y$ -axes the standard equation becomes

$\frac{{(x - α)}^{2}}{a^{2}} + \frac{{(y - β)}^{2}}{b^{2}} = 1.$

Task!

Consider the points $A$ and $B$ with Cartesian coordinates $(c, 0)$ and $(- c, 0)$ respectively. A curve has the property that for every point $P$ on it the sum of the distances $P A$ and $P B$ is a constant (which we will call $2 a$ ). Derive the Cartesian form of the equation of the curve and show that it is an ellipse.

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We use Pythagoras’s theorem to work out the distances $P A$ and $P B$ :

$Let R_{1} = P B = {[{(x + c)}^{2} + y^{2}]}^{1 ∕ 2}$ and let $R_{2} = P A = {[{(c - x)}^{2} + y^{2}]}^{1 ∕ 2}$

We now take the given equation $R_{1} + R_{2} = 2 a$ and multiply both sides by $R_{1} - R_{2}$ . The quantity $R_{1}^{2} - R_{2}^{2}$ on the left is calculated to be $4 c x$ , and $2 a (R_{1} - R_{2})$ is on the right. We thus obtain a pair of equations: $R_{1} + R_{2} = 2 a$ and $R_{1} - R_{2} = \frac{2 c x}{a}$

Adding these equations together gives $R_{1} = a + \frac{c x}{a}$ and squaring this equation gives

$x^{2} + c^{2} + 2 c x + y^{2} = a^{2} + \frac{c^{2} x^{2}}{a^{2}} + 2 c x$

Simplifying: $x^{2} (1 - \frac{c^{2}}{a^{2}}) + y^{2} = a^{2} - c^{2}$ whence $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{(a^{2} - c^{2})} = 1$

This is the standard equation of an ellipse if we set $b^{2} = a^{2} - c^{2}$ , which is the traditional equation which relates the two semi-axis lengths $a$ and $b$ to the distance $c$ of the foci from the centre of the ellipse.

The foci $A$ and $B$ have optical properties; a beam of light travelling from $A$ along $A P$ and undergoing a mirror reflection from the ellipse at $P$ will return along the path $P B$ to the other focus $B$ .

1.2 The circle

The circle is a special case of the ellipse; it occurs when $a = b = r$ so the equation becomes

$\frac{x^{2}}{r^{2}} + \frac{y^{2}}{r^{2}} = 1 or, more commonly x^{2} + y^{2} = r^{2}$

Here, the centre of the circle is located at the origin $(0, 0)$ and the radius of the circle is $r$ . If the centre of the circle at a point $(α, β)$ then the equation takes the form:

${(x - α)}^{2} + {(y - β)}^{2} = r^{2}$

Key Point 2

The equation of a circle with centre at $(α, β)$ and radius $r$ is ${(x - α)}^{2} + {(y - β)}^{2} = r^{2}$

Task!

Write down the equations of the five circles ( $A$ to $E$ ) below:

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$A$	${(x - 1)}^{2} + {(y - 1)}^{2} = 1$

$B$	${(x - 3)}^{2} + {(y - 1)}^{2} = 1$

$C$	${(x + 0.5)}^{2} + {(y + 2)}^{2} = 1$

$D$	${(x - 2)}^{2} + {(y + 2)}^{2} = {(0.5)}^{2}$

$E$	${(x + 0.5)}^{2} + {(y - 2.5)}^{2} = 1$

Example 2

Show that the expression

$x^{2} + y^{2} - 2 x + 6 y + 6 = 0$

represents the equation of a circle. Find its centre and radius.

Solution

We shall see later how to recognise this as the equation of a circle simply by examination of the coefficients of the quadratic terms $x^{2}, y^{2}$ and $x y$ . However, in the present example we will use the process of completing the square, for $x$ and for $y$ , to show that the expression can be written in standard form.

Now $x^{2} + y^{2} - 2 x + 6 y + 6 \equiv x^{2} - 2 x + y^{2} + 6 y + 6$ .

Also,

$x^{2} - 2 x \equiv {(x - 1)}^{2} - 1 and y^{2} + 6 y \equiv {(y + 3)}^{2} - 9.$

Hence we can write

$x^{2} + y^{2} - 2 x + 6 y + 6 \equiv {(x - 1)}^{2} - 1 + {(y + 3)}^{2} - 9 + 6 = 0$

or, taking the free constants to the right-hand side:

${(x - 1)}^{2} + {(y + 3)}^{2} = 4.$

By comparing this with the standard form we conclude this represents the equation of a circle with centre at $(1, - 3)$ and radius 2.

Task!

Find the centre and radius of each of the following circles:

$x^{2} + y^{2} - 4 x - 6 y = - 12$
$2 x^{2} + 2 y^{2} + 4 x + 1 = 0$

centre: $(2, 3)$ radius 1
centre: $(- 1, 0)$ radius $\sqrt{2} ∕ 2.$

1 The ellipse, parabola and hyperbola

1.1 The ellipse

Answer

1.2 The circle

Answer

Answer