HSC Extension 1 and 2 Mathematics/3-Unit/Preliminary/Parametrics

The parametric form of a curve is an algebraic representation which expresses the co-ordinates of each point on the curve as a function of an introduced parameter, most frequently $t$. This contrasts with Cartesian form in that parametric equations do not describe an explicit relation between $x$ and $y$. This relation must be derived in order to convert from parametric to Cartesian form.

In 3-unit, parametrics focuses on a parametric representation of the quadratic, moving from parametric to Cartesian form and vice-versa, and manipulating geometrical aspects of the quadratic with parametrics. Recognition of other parametric forms is also useful, and more forms are introduced and dealt with in the 4-unit topic, [[../../../4-Unit/Conics|conics]].

Reasons for using the parametric form

In specific situations (in the school syllabus, primarily Conic sections), the parametric representation can be useful because:

• points on the curve are represented by a single number, not two, simplifying algebra;
• some elegant results are possible; for instance, in the standard paramaterisation of the quadratic, the gradient is equal to the parameter, $t$;
• some curves, which cannot be expressed in functional form (for example, the circle, which is neither a function of $x$ nor of $y$) can be conveniently expressed in parametric form;
• intuitively, it allows for an easier way to find points on the graph: you can sub in any value for the parameter and instantly find a point, whereas a relational form is not deterministic in the same way.

Furthermore, the parametric form occurs in certain natural phenomena. For instance, using the equations of motion, a thrown ball's location at any time can be calculated using the laws of projectile motion. This is implicitly paramaterising the ball's path by time; to find the shape of the ball's path, (which we know is a parabola) we must use parametrics, to eliminate time, $t$, from the equations.

Converting parametric to Cartesian forms

One of the simplest parametric forms is the line:

$\begin{array}{cc}x& =t\\ y& =t\end{array}$

By inspection, it is obvious that this describes the line $y=x$. However, what is the formalised approach for doing this?

We are looking for some relation between $x$ and $y$ which has no $t$ in it. In other words, we want to eliminate $t$ from the equations. In the above example, we did this by equating the first and second equations, eliminating $t$.

Another example:

$\begin{array}{cc}x& =3t+1\\ y& =9t^2-1\end{array}$

We solve equation 1 for t, and substitute into equation 2:

$\begin{array}{cc}x& =3t+1\\ t& =\frac{x-1}{3}\end{array}$

Sub into equation 2:

$\begin{array}{cc}y& =9t^2-1\\ & =9{\left(\frac{x-1}{3}\right)}^2-1\\ & =x^2-2x+1-1\\ & =x^2-2x\end{array}$

Here we have a Cartesian form, as required.

Three standard parametric forms

These parametric forms occur frequently, and should be recognised by 3- and 4-unit students.

Parabola

The standard paramaterisation of the parabola describes one with focal length $a$ and with its vertex at the origin. In Cartesian form, this is given as

$x^2=4ay$

In parametric form, this is given as

$\begin{array}{cc}x& =2at\\ y& =a{t}^2\end{array}$

Eliminating $t$ allows us to verify that this is equivalent to the Cartesian form:

$\begin{array}{cc}x& =2at\\ x^2& =4a^2{t}^2\\ & =4a(a{t}^2)\\ & =4ay\end{array}$

as required.

Circle

Diagram of a circle, indicating the geometrical meaning of the parameter

The circle with radius $r$ and center at the origin can be written in Cartesian form as

$x^2+y^2=r^2$

Introducing the parameter, $\theta$, this is:

$\begin{array}{cc}x& =r\cos \theta \\ y& =r\sin \theta \end{array}$

We convert to Cartesian as follows:

$\begin{array}{cc}{x}^2& =r^2{\cos }^2\theta \\ y^2& =r^2{\sin }^2\theta \\ x^2+y^2& =r^2{\cos }^2\theta +r^2{\sin }^2\theta \\ & =r^2({\cos }^2\theta +{\sin }^2\theta )\end{array}$

Recalling the trig identity,

${\sin }^2\theta +{\cos }^2\theta =1$

we conclude

$x^2+y^2=r^2$

as required.

Geometrical interpretation

Unlike most parametrics, the conic sections are paramaterised by $\theta$ instead of $t$. For the circle, this implies a geometric representation: $\theta$ represents the angle the point makes with the $x$ axis.

Ellipse

Conceptually, an ellipse is just a 'squished' circle. The parametric form makes this clear:

$\begin{array}{cc}x& =a\cos \theta \\ y& =b\sin \theta \end{array}$

The cos and sin are still there, but they are now multiplied by different constants so that the $x$ and $y$ components are stretched differently. We can turn this into the parametric form in a similar fashion to the circle:

$\begin{array}{cc}{x}^2& =a^2{\cos }^2\theta \\ y^2& =b^2{\sin }^2\theta \\ \frac{x^2}{a^2}+\frac{y^2}{b^2}& ={\cos }^2\theta +{\sin }^2\theta \\ & =1\end{array}$

which is the standard form of an ellipse, with $x$ and $y$ intercepts at $\pm a$ and $\pm b$, respectively.

Properties of the parabola

3-Unit students are expected to remember the parametric description of a parabola ($x=2at,y=a{t}^2$). They are also expected to know (and/or be able to quickly derive) the equations of the tangent and the normal to the parabola at the point $t$ and at the point $(x_1,y_1)$.

Derivation of equations of the tangent and the normal

Differentiating each parametric equation with respect to $t$,

$\begin{array}{cc}\frac{dy}{dt}& =2at\\ \frac{dx}{dt}& =2a\end{array}$

Then the gradient $\frac{dy}{dx}$ can be obtained by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$ (this is the chain rule):

$\begin{array}{cc}\frac{dy}{dx}& =\frac{dy}{dt}\frac{dt}{dx}\\ & =\frac{2at}{1}\frac{1}{2a}\\ & =t\end{array}$

Note that this result could have been derived without the chain rule, by taking the derivative of the Cartesian form (with respect to $x$) and solving for $t$. However, the above derivation is faster and more elegant.

The equation of the tangent at P

Diagram of a parabola, showing the normal and tangent at a point P

The gradient of the tangent at a point $P(2at,a{t}^2)$ then is $t$. The equation of the tangent at $P$ is:

Using the point-gradient formula $y-y_1=m(x-x_1)$, the equation is:

$\begin{array}{cc}y-a{t}^2& =t(x-2at)\\ & =tx-2a{t}^2\\ y+a{t}^2& =tx\\ y-tx+a{t}^2& =0& \text{ or }y=xt-a{t}^2\end{array}$

The equation of the normal at P

The gradient of the normal at the point $P$ is $\frac{-1}{t}$ (since two perpendicular lines with gradients $m_1$ and $m_2$ must have $m_1{m}_2=-1$). Similarly to the derivation of the equation of the tangent:

$\begin{array}{cc}y-a{t}^2& =\frac{-1}{t}(x-2at)\\ & =\frac{-x}{t}+\frac{2at}{t}\\ yt-a{t}^3& =-x+2at\\ yt+x-at(t^2+2)& =0& \text{or }x+ty=2at+a{t}^3\end{array}$

Chords of a parabola

Diagram of a parabola, a chord PQ

Suppose that $P(2ap,a{p}^2)$ and $Q(2aq,a{q}^2)$ are two distinct points on the parabola $x^2=4ay$. We can derive the equation for the line $PQ$ by finding the gradient and using the point-gradient formula:

$\begin{array}{cc}m& =\frac{a{p}^2-a{q}^2}{2ap-2aq}\\ & =\frac{a(p-q)(p+q)}{2a(p-q)}\\ & =\frac{1}{2}(p+q)\end{array}$

so the chord is

$\begin{array}{cc}y-a{p}^2& =\frac{1}{2}(p+q)(x-2ap)\\ & =\frac{1}{2}(p+q)x-a{p}^2-apq\\ y& =\frac{1}{2}(p+q)x-apq\end{array}$

Intersection of tangents

The points $P(2ap,a{p}^2)$ and $Q(2aq,a{q}^2)$ lie on the parabola $x^2=4ay$. The intersection of the tangents at $P$ and $Q$ can be found in terms of $p$ and $q$:

The tangents are:

$\begin{array}{cc}y& =px-a{p}^2\\ y& =qx-a{q}^2\end{array}$

Subtracting these,

$\begin{array}{cc}y-y& =px-a{p}^2-(qx-a{q}^2)\\ 0& =px-a{p}^2-qx+a{q}^2\\ 0& =px-qx+a{q}^2-a{p}^2\\ x(p-q)& =a(p^2-q^2)\\ x(p-q)& =a(p+q)(p-q)\\ x& =a(p+q)\end{array}$

Substituting into the original tangent formula,

$\begin{array}{cc}y& =p(a(p+q))-a{p}^2\\ & =ap(p+q)-a{p}^2\\ & =a{p}^2+apq-a{p}^2\\ & =apq\end{array}$

So the point of intersection is described by $T(a(p+q),apq)$.