# 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, 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:
    ${\begin{aligned}x&=t\\y&=t\end{aligned}}$
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:

    ${\begin{aligned}x&=3t+1\\y&=9t^{2}-1\end{aligned}}$
We solve equation 1 for t, and substitute into equation 2:
    ${\begin{aligned}x&=3t+1\\t&={\frac {x-1}{3}}\end{aligned}}$
Sub into equation 2:
    ${\begin{aligned}y&=9t^{2}-1\\&=9\left({\frac {x-1}{3}}\right)^{2}-1\\&=x^{2}-2x+1-1\\&=x^{2}-2x\end{aligned}}$
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
    ${\begin{aligned}x&=2at\\y&=at^{2}\end{aligned}}$
Eliminating $t\,$ allows us to verify that this is equivalent to the Cartesian form:
    ${\begin{aligned}x&=2at\\x^{2}&=4a^{2}t^{2}\\&=4a(at^{2})\\&=4ay\end{aligned}}$
as required.

### Circle

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:
    ${\begin{aligned}x&=r\cos \theta \\y&=r\sin \theta \end{aligned}}$
We convert to Cartesian as follows:
    ${\begin{aligned}x^{2}&=r^{2}\cos ^{2}\theta \\y^{2}&=r^{2}\sin ^{2}\theta \\x^{2}+y^{2}&=r^{2}\cos ^{2}\theta +r^{2}\sin ^{2}\theta \\&=r^{2}(\cos ^{2}\theta +\sin ^{2}\theta )\end{aligned}}$
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:
    ${\begin{aligned}x&=a\cos \theta \\y&=b\sin \theta \end{aligned}}$
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:
    ${\begin{aligned}x^{2}&=a^{2}\cos ^{2}\theta \\y^{2}&=b^{2}\sin ^{2}\theta \\{\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}&=\cos ^{2}\theta +\sin ^{2}\theta \\&=1\end{aligned}}$
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 ($x=2at,y=at^{2}$). 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

Differentiating each parametric equation with respect to $t\,$,
    ${\begin{aligned}{\frac {dy}{dt}}&=2at\\{\frac {dx}{dt}}&=2a\end{aligned}}$
Then the gradient ${\tfrac {dy}{dx}}$ can be obtained by dividing ${\tfrac {dy}{dt}}$ by ${\tfrac {dx}{dt}}$ (this is the chain rule):
    ${\begin{aligned}{\frac {dy}{dx}}&={\frac {dy}{dt}}{\frac {dt}{dx}}\\&={\frac {2at}{1}}{\frac {1}{2a}}\\&=t\end{aligned}}$
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

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:
       ${\begin{aligned}y-at^{2}&=t(x-2at)\\&=tx-2at^{2}\\y+at^{2}&=tx\\y-tx+at^{2}&=0&{\mbox{ or }}y=xt-at^{2}\end{aligned}}$

#### The equation of the normal at P

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

### Chords of a parabola

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:
    ${\begin{aligned}m&={\frac {ap^{2}-aq^{2}}{2ap-2aq}}\\&={\frac {a(p-q)(p+q)}{2a(p-q)}}\\&={\tfrac {1}{2}}(p+q)\end{aligned}}$
so the chord is
    ${\begin{aligned}y-ap^{2}&={\tfrac {1}{2}}(p+q)(x-2ap)\\&={\tfrac {1}{2}}(p+q)x-ap^{2}-apq\\y&={\tfrac {1}{2}}(p+q)x-apq\end{aligned}}$

### Intersection of tangents

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:
       ${\begin{aligned}y&=px-ap^{2}\\y&=qx-aq^{2}\end{aligned}}$
    Subtracting these,
       ${\begin{aligned}y-y&=px-ap^{2}-(qx-aq^{2})\\0&=px-ap^{2}-qx+aq^{2}\\0&=px-qx+aq^{2}-ap^{2}\\x(p-q)&=a(p^{2}-q^{2})\\x(p-q)&=a(p+q)(p-q)\\x&=a(p+q)\end{aligned}}$
    Substituting into the original tangent formula,
       ${\begin{aligned}y&=p(a(p+q))-ap^{2}\\&=ap(p+q)-ap^{2}\\&=ap^{2}+apq-ap^{2}\\&=apq\end{aligned}}$
So the point of intersection is described by $T(a(p+q),apq)\,$.