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

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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, conics.

Reasons for using the parametric form[edit]

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[edit]

One of the simplest parametric forms is the line:

x & = t \\
y & = t\end{align}

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:[edit]

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

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

    x & = 3t + 1 \\
    t & = \frac{x - 1}{3}

Sub into equation 2:

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

Here we have a Cartesian form, as required.

Three standard parametric forms[edit]

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


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{align}x & = 2at \\
y & = at^2\end{align}

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

    x   & = 2at \\
    x^2 & = 4a^2t^2 \\
        & = 4a (at^2) \\
        & = 4ay

as required.


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{align}x &= r\cos\theta \\
y &= r\sin\theta\end{align}

We convert to Cartesian as follows:

    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)

Recalling the trig identity,

\sin^2\theta + \cos^2\theta = 1\,

we conclude

x^2 + y^2 = r^2\,

as required.

Geometrical interpretation[edit]

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.


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

\begin{align}x &= a\cos\theta\\
y &= b\sin\theta\end{align}

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:

    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

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

Properties of the parabola[edit]

3-Unit students are expected to remember the parametric description of a parabola (x = 2at, y = at^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[edit]

Differentiating each parametric equation with respect to t\,,

\frac{dy}{dt} & = 2at \\
\frac{dx}{dt} & = 2a

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

\frac{dy}{dx} & = \frac{dy}{dt} \frac{dt}{dx} \\
              & = \frac{2at}{1} \frac{1}{2a} \\
              & = t

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[edit]

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

The gradient of the tangent at a point P(2at,at^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:
y-at^2 & = t(x-2at) \\
       & = tx - 2at^2 \\
y+at^2 & = tx \\
y-tx+at^2 & = 0 & \mbox{ or } y = xt - at^2

The equation of the normal at P[edit]

The gradient of the normal at the point P is \tfrac{-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:

y - at^2 & = \tfrac{-1}{t}(x - 2at) \\
         & = \tfrac{-x}{t} + \tfrac{2at}{t} \\
yt - at^3 & = -x + 2at \\
yt + x - at(t^2 + 2) & = 0 & \mbox{or } x + ty = 2at + at^3

Chords of a parabola[edit]

Diagram of a parabola, a chord PQ

Suppose that P(2ap,ap^2) and Q(2aq,aq^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:

m &= \frac{ap^2-aq^2}{2ap-2aq} \\
  &= \frac{a(p-q)(p+q)}{2a(p-q)} \\
  &= \tfrac{1}{2}(p+q)

so the chord is

y - ap^2 &= \tfrac{1}{2}(p+q)(x-2ap) \\
         &= \tfrac{1}{2}(p+q)x - ap^2 - apq \\
       y &= \tfrac{1}{2}(p+q)x - apq

Intersection of tangents[edit]

The points P(2ap,ap^2) and Q(2aq,aq^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:
y & = px - ap^2 \\
y & = qx - aq^2
Subtracting these,
y - y & = px - ap^2 - (qx - aq^2) \\
0 & = px - ap^2 - qx + aq^2 \\
0 & = px - qx + aq^2 - ap^2 \\
x(p-q) & = a(p^2 - q^2) \\
x(p-q) & = a(p+q)(p-q) \\
x & = a(p+q)
Substituting into the original tangent formula,
y & = p(a(p+q)) - ap^2 \\
  & = ap(p+q) - ap^2 \\
  & = ap^2 + apq - ap^2 \\
  & = apq

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