3 Standard forms of conic sections in parametric form

We have seen above that, given a curve in the $xy$ plane, there is no unique way of representing it in parametric form. However, for some commonly occurring curves, particularly the conics, there are accepted standard parametric equations.

3.1 The parabola

The standard parametric equations for a parabola are: $x=a{t}^{2}y=2at$

Clearly, we have  $t=\frac{y}{2a}$  and by eliminating $t$ we get $x=a\left(\frac{y^2}{4a^2}\right)$   or   $y^2=4ax$ which we recognise as the standard Cartesian description of a parabola. As an illustration, Figure 19 shows the curve with $a=2$ and $-1\le t\le 2.3$

Figure 19

3.2 The ellipse

Here, the standard equations are    $x=acosty=bsint$

Again, eliminating $t$ (dividing the first equation by $a$ , the second by $b$ , squaring and adding) we have

${\left(\frac{x}{a}\right)}^2+{\left(\frac{y}{b}\right)}^2={cos}^{2}t+{sin}^{2}t\equiv 1$   or, in more familiar form: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1.$

If we choose the range for $t$ as $0\le t\le \frac{7\pi }{4}$ the following segment of the ellipse is obtained.

Figure 20

Here we note that (except in the special case when $a=b$ , giving a circle) the parameter $t$ is not the angle that the radial line makes with the the positive $x$ -axis. In the study of the orbits of planets and satellites it is often preferable to use plane polar coordinates $\left(r,\theta \right)$ to treat the problem. In these coordinates an ellipse has an equation of the form $\frac{1}{r}=A+Bcos\theta$ , with $A$ and $B$ positive numbers such that $B<A$ . Not only is there a difference in the equations on passing from Cartesian to polar coordinates; there is also a change in the origin of coordinates. The polar coordinate equation is using a focus of the ellipse as the origin. In the Cartesian description the foci are two points at $+e$ along the $x$ -axis, where $e$ obeys the equation $e=a-b$ , if we assume that $a<b$ i.e. we choose the long axis of the ellipse as the $x$ -axis. This problem gives some practice at algebraic manipulation and also indicates some shortcuts which can be made once the mathematics of the ellipse has been understood.

Example 9

An ellipse is described in plane polar coordinates by the equation

$\frac{1}{r}=2+cos\theta$

Convert the equation to Cartesian form. [Hint: remember that $x=rcos\theta$ .]

Solution

Multiplying the given equation by $r$ and then using $x=rcos\theta$ gives the results

$1=2r+x$ so that $2r=1-x$

We now square the second equation, remembering that $r^2=x^2+y^2$ . We now have

$4\left(x^2+y^2\right)={\left(1-x\right)}^2=1+x^2-2x$ so that $3x^2+2x+4y^2=1$

We now recall the method of completing the square, which allows us to set

$3x^2+2x=3{\left(x^2+\frac{2x}{3}\right)}^2-\frac{1}{9}$

Putting this result into the equation and collecting terms leads to the final result

$\frac{{\left(x+\frac{1}{3}\right)}^2}{a^2}+\frac{y^2}{b^2}=1$ with $a=\frac{2}{3}$ and $b=\sqrt{\frac{1}{3}}$ .

This is the standard Cartesian form for the equation of an ellipse but we must remember that we started from a polar equation with a focus of the ellipse as origin. The presence of the term $x+\frac{1}{3}$ in the equation above actually tells us that the focus being used as origin was a distance of $\frac{1}{3}$ to the right of the centre of the ellipse at $x=0$ .

The preceding piece of algebra was necessary in order to convince us that the original equation in plane polar coordinates does represent an ellipse. However, now that we are convinced of this we can go back and try to extract information in a more speedy way from the equation in its original $\left(r,\theta \right)$ form.

Try setting $\theta =0$ and $\theta =\pi$ in the equation

$\frac{1}{r}=2+cos\theta$

We find that at $\theta =0$ we have $r=\frac{1}{3}$ while at $\theta =\pi$ we have $r=1$ . These $r$ values correspond to the two ends of the ellipse, so the long axis has a total length $1+\frac{1}{3}=\frac{4}{3}$ . This tells us that $a=\frac{2}{3}$ , exactly as found by our longer algebraic derivation. We can further deduce that the focus acting as origin must be at a distance of $\frac{1}{3}$ from the centre of the ellipse in order to lead to the two $r$ values at $\theta =0$ and $\theta =\pi$ . If we now use the equation $e=a-b$ mentioned earlier then we find that $\frac{1}{9}=\frac{4}{9}-b^2$ , so that $b=\sqrt{\frac{1}{3}}$ , as obtained by our lengthy algebra.

3.3 The hyperbola

The standard equations are    $x=acoshty=bsinht$ .

In this case, to eliminate $t$ we use the identity ${cosh}^{2}t-{sinh}^{2}t\equiv 1$ giving rise to the equation of the hyperbola in Cartesian form:

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

In Figure 21 we have chosen a parameter range $-1\le t\le 2$ .

To obtain the complete curve the parameter range $-\infty <t<\infty$ must be used. These parametric equations only give the right-hand branch of the hyperbola. To obtain the left-hand branch we would use $x=-acoshty=bsinht$

Exercises

1. In the following sketch the given parametric curves. Also, eliminate the parameter to give the Cartesian equation in $x$ and $y$ .
   1. $x=t,y=2-t0\le t\le 1$
   2. $x=2-t,y=t+10\le t\le \infty$
   3. $x=\frac{2}{t}y=t-20<t<3$  
   4. $x=3sin\frac{\pi t}{2}y=4cos\frac{\pi t}{2}-1\le t\le 0.5$
2. Find the tangent line to the parametric curve $x=t^2-t$ $y-t^2+t$ at the point where $t=1$ .
3. For each of the following curves expressed in parametric form obtain expressions for $\frac{dy}{dx}$ and $\frac{d^{2}y}{d{x}^2}$ and use this information to help make a sketch.
   1. $x=t^2-2t,y=t^2-4t$
   2. $x=t^3-3t-2,y=t^2-t-2$

Answer