# 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 ${\displaystyle t}$. This contrasts with Cartesian form in that parametric equations do not describe an explicit relation between ${\displaystyle x}$ and ${\displaystyle y}$. 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, ${\displaystyle t}$;
• some curves, which cannot be expressed in functional form (for example, the circle, which is neither a function of ${\displaystyle x}$ nor of ${\displaystyle y}$) 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, ${\displaystyle t}$, from the equations.

## Converting parametric to Cartesian forms

One of the simplest parametric forms is the line:

{\displaystyle {\begin{aligned}x&=t\\y&=t\end{aligned}}}

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

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

#### Another example:

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

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

{\displaystyle {\begin{aligned}x&=3t+1\\t&={\frac {x-1}{3}}\end{aligned}}}

Sub into equation 2:

{\displaystyle {\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 ${\displaystyle a\,}$ and with its vertex at the origin. In Cartesian form, this is given as

${\displaystyle x^{2}=4ay\,}$

In parametric form, this is given as

{\displaystyle {\begin{aligned}x&=2at\\y&=at^{2}\end{aligned}}}

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

{\displaystyle {\begin{aligned}x&=2at\\x^{2}&=4a^{2}t^{2}\\&=4a(at^{2})\\&=4ay\end{aligned}}}

as required.

### Circle

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

The circle with radius ${\displaystyle r\,}$ and center at the origin can be written in Cartesian form as

${\displaystyle x^{2}+y^{2}=r^{2}\,}$

Introducing the parameter, ${\displaystyle \theta \,}$, this is:

{\displaystyle {\begin{aligned}x&=r\cos \theta \\y&=r\sin \theta \end{aligned}}}

We convert to Cartesian as follows:

{\displaystyle {\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,

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

we conclude

${\displaystyle x^{2}+y^{2}=r^{2}\,}$

as required.

#### Geometrical interpretation

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

### Ellipse

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

{\displaystyle {\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 ${\displaystyle x}$ and ${\displaystyle y}$ components are stretched differently. We can turn this into the parametric form in a similar fashion to the circle:

{\displaystyle {\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 ${\displaystyle x}$ and ${\displaystyle y}$ intercepts at ${\displaystyle \pm a\,}$ and ${\displaystyle \pm b\,}$, respectively.

## Properties of the parabola

3-Unit students are expected to remember the parametric description of a parabola (${\displaystyle x=2at,y=at^{2}}$). 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 ${\displaystyle t}$ and at the point ${\displaystyle (x_{1},y_{1})}$.

### Derivation of equations of the tangent and the normal

Differentiating each parametric equation with respect to ${\displaystyle t\,}$,

{\displaystyle {\begin{aligned}{\frac {dy}{dt}}&=2at\\{\frac {dx}{dt}}&=2a\end{aligned}}}

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

{\displaystyle {\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 ${\displaystyle x\,}$) and solving for ${\displaystyle t\,}$. 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 ${\displaystyle P(2at,at^{2})}$ then is ${\displaystyle t}$. The equation of the tangent at ${\displaystyle P}$ is:

Using the point-gradient formula ${\displaystyle y-y_{1}=m(x-x_{1})\,}$, the equation is:
{\displaystyle {\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 ${\displaystyle P}$ is ${\displaystyle {\tfrac {-1}{t}}}$ (since two perpendicular lines with gradients ${\displaystyle m_{1}\,}$ and ${\displaystyle m_{2}\,}$ must have ${\displaystyle m_{1}m_{2}=-1\,}$). Similarly to the derivation of the equation of the tangent:

{\displaystyle {\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

Diagram of a parabola, a chord PQ

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

{\displaystyle {\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

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

The tangents are:
{\displaystyle {\begin{aligned}y&=px-ap^{2}\\y&=qx-aq^{2}\end{aligned}}}
Subtracting these,
{\displaystyle {\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,
{\displaystyle {\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 ${\displaystyle T(a(p+q),apq)\,}$.