# HSC Mathematics Extension 1 – Parametric Equations

Parametric Equations

## Parametric Representation

At times it is useful to express two related variables, such as $Latex formula$ and $Latex formula$, in terms of a third variable, $Latex formula$. Doing so gives,

$Latex formula$

$Latex formula$

These equations are called parametric equations, and $Latex formula$ is called a parameter. Indeed, this technique becomes increasingly important upon studying the algebraic representation of vectors.

It is particularly useful to express the equation of the parabola $Latex formula$ in terms of two parametric equations. These equations are $Latex formula$, $Latex formula$. By the simple elimination of $Latex formula$ we obtain the general Cartesian equation expressed above. The parabola and its parametric equations form the focus of this topic.

As mentioned above, the parametric equations of the parabola $Latex formula$ are given by:

 $Latex formula$ $Latex formula$

It should be noted that the sign of $Latex formula$ indicates whether or not the parabola is concave up or down. If $Latex formula$ then it follows that the parabola is concave up, and if $Latex formula$ then it follows that the parabola is concave down. Also, as a side note $Latex formula$, the absolute value of the value of $Latex formula$, gives the focal length.

## Converting From Parametric Form to Cartesian form and vice versa

It is quite simple to express a parabola in parametric form. To do so simply find the corresponding value of $Latex formula$ or the parabola with equation $Latex formula$ and then substitute into the equations,

$Latex formula$

$Latex formula$

To convert a parabola from a set of parametric equations to a Cartesian equation, simply eliminate the value of $Latex formula$ by simple substitution. For example, to obtain the Cartesian equation of the parabola with parametric equations

$Latex formula$

$Latex formula$

We would simply perform the following algebraic manipulations,

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

Below is an example which illustrates the process of converting from Cartesian form to parametric form.

### Example 1

Find the Parametric equations of the parabola $Latex formula$.

### Solution 1

In this example, we simply find the value $Latex formula$ and substitute into the standard parametric equations. So we have that,

$Latex formula$

Hence, upon comparing the above equation to the general form $Latex formula$, we obtain the value of $Latex formula$ which is $Latex formula$. Hence the parametric equations are

$Latex formula$

$Latex formula$

Here is an example where the parametric equations are converted into a single Cartesian equation.

### Example 2

Find the Cartesian equation of the parabola with Parametric equations $Latex formula$, $Latex formula$.

### Solution 2

We shall simply eliminate the value of $Latex formula$ from the two equations to obtain a Cartesian equation.

$Latex formula$

Hence, upon substitution we obtain,

$Latex formula$

$Latex formula$

$Latex formula$

which is the Cartesian equation.

Here is a slightly more difficult question that requires some thought.

### Example 3

Find the parametric equations of the parabola $Latex formula$.

### Solution 3

Firstly we rearrange the equation to obtain the value of $Latex formula$.

$Latex formula$

Hence the value of $Latex formula$ is $Latex formula$.

Now, although this is omitted in most texts, the actual parametric equations of the parabola $Latex formula$ where $Latex formula$ is any non-zero number are, $Latex formula$. So in this case, we have that the parametric equations are,

$Latex formula$

$Latex formula$

We shall now investigate the equations of the chord, tangent and normal to the parabola $Latex formula$.

## Tangents, Normals and Chords

It should be noted that, upon studying this topic students should feel comfortable with all the derivation processes of the tangents, normals and chords in both parametric and Cartesian form. It is not particularly important to remember the equations of each of these, rather, it is important that the student be able to realise the relationships between the differing anatomical features of the parabola.

### Tangents

We shall firstly investigate the equation of the tangent at the point $Latex formula$ lying on the parabola $Latex formula$. Note that the tangent found with such coordinates is called the Cartesian form of the tangent. Conversely, the equation of the tangent found at the point $Latex formula$ is dubbed the Parametric form of the tangent.

The Cartesian equation of the tangent at the point $Latex formula$ on the parabola is $Latex formula$,

 $Latex formula$

The parametric form of the equation of the tangent at the point $Latex formula$ is,

 $Latex formula$

### Normals

The Cartesian equation of the normal at the point $Latex formula$ on the parabola $Latex formula$ is,

 $Latex formula$

The parametric form of the equation of the normal at the point $Latex formula$ is,

 $Latex formula$

### Chords

The equation of the chord between the points $Latex formula$ and $Latex formula$ on the parabola $Latex formula$ is

$Latex formula$

Now, suppose that the chord is a focal chord. That is, the chord passes through the focus $Latex formula$ of the parabola. Then we have that

That is,

$Latex formula$

It is also obvious that if $Latex formula$, then the chord $Latex formula$ passes through the focus $Latex formula$.

Hence we have that a necessary and sufficient condition for a chord to be a focal chord is that the parameters at their respective points must multiply to equal $Latex formula$. This fact becomes important later on.

### Chord of Contact

The chord of contact to the parabola $Latex formula$ from the external point $Latex formula$ is the chord that joins the two points from which tangents are drawn from the parabola that pass through the point $Latex formula$. Consider the diagram below. Here, $Latex formula$ is the chord of contact from the point $Latex formula$.

To show that a point $Latex formula$ is external to the parabola, we simply substitute the point into the equation $Latex formula$, and if $Latex formula$ then the point lies external to the parabola. Otherwise, the point lies on or within the parabola and hence is not external to the parabola.

The equation of the chord of contact from the external point $Latex formula$ to the parabola $Latex formula$ has equation $Latex formula$. Notice the similarity to the equation of the tangent. The proof of this is remarkably elegant, and explains why there is such a great similarity between general equation of the tangent and the equation of the chord of contact.

The equation of the chord of contact from the external point $Latex formula$ on the parabola $Latex formula$ is,

 $Latex formula$

We shall now consider some examples.

### Example 4

Find the equations of the tangents to the parabola $Latex formula$ at the points $Latex formula$ and $Latex formula$. Find also the point of intersection of the two tangents.

### Solution 4

In this question, we shall not derive the equation but shall illustrate use of the formulae derived. Note however, that students should always derive the equations accordingly, instead of remembering formulae. So, comparing the parametric form of this parabola with the standard parametric form gives, $Latex formula$. Hence we have that the equation of the tangent at the point $Latex formula$ where coordinates are given by:

$Latex formula$

$Latex formula$

When $Latex formula$ we have the equation of the tangent being,

$Latex formula$

$Latex formula$

Now, solving simultaneously to find the point of intersection gives,

$Latex formula$

Hence the two tangents intersect at the point $Latex formula$

### Example 5

The straight line $Latex formula$ cuts the parabola $Latex formula$ at the points $Latex formula$ and $Latex formula$. Find:

a) The coordinates of $Latex formula$ and $Latex formula$

b) The equations of the normals at $Latex formula$ and $Latex formula$

c) Show that the normals intersect on the parabola.

### Solution 5

a) To solve this section, we must solve the equations of the parabola and the straight line simultaneously. Doing so gives,

$Latex formula$

$Latex formula$

$Latex formula$

Hence the coordinates of $Latex formula$ and $Latex formula$ are,

$Latex formula$ and $Latex formula$

b) Students at this point must derive the formulae using first principles. However, the derivation of the formulae is exactly the same as that done for the derived formulae, and hence we shall use the derived formulae. Hence we obtain the normal at $Latex formula$:

$Latex formula$

$Latex formula$

$Latex formula$

Also, the normal at $Latex formula$ has equation $Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

c) To find the point of intersection of the parabolas, it is necessary to solve their equations algebraically. So we have,

$Latex formula$

$Latex formula$

$Latex formula$ gives,

$Latex formula$

And hence we have that $Latex formula$. Thus the point of intersection of the normals is $Latex formula$.

### Example 6

Find the equation of the chord joining $Latex formula$ and $Latex formula$ on $Latex formula$ and $Latex formula$:

a) By first principles;

b) By using the formulae already derived.

c) Also show that is a focal chord.

### Solution 6

a) To find the equation by first principles, we simply find the points corresponding to the parameters $Latex formula$ and $Latex formula$.

When $Latex formula$, $Latex formula$. Hence the point is $Latex formula$.

When$Latex formula$ Hence the point is:

$Latex formula$

Now, we must find the gradient. Doing so gives,

$Latex formula$

Now, using the point gradient formula,

$Latex formula$

Hence the equation of $Latex formula$ is

$Latex formula$

b) Now, the formulae derived is

$Latex formula$

Now, let $Latex formula$. Also, upon comparing the parametric form with the standard parametric form of the parabola, we have that $Latex formula$. Thus, using this we obtain:

$Latex formula$

$Latex formula$

That is,

$Latex formula$

c) To show that it is a focal chord, we shall simply show that the focus $Latex formula$ passes through the chord. Now, since $Latex formula$, then it follows that the focus has coordinates $Latex formula$.

Now, substituting into the equation of the chord gives:

$Latex formula$

Hence the chord passes through the focus and is a focal chord.

Note that an alternative method is to show that product of the parameters for the endpoints of the chord has a product of $Latex formula$. That is, $Latex formula$. Recall that this was found to be a necessary and sufficient condition for the chord $Latex formula$ to be a focal chord.

### Example 7

Show that the point $Latex formula$ is external to the parabola $Latex formula$. Hence,

a) Find the equation of the chord of contact to the parabola $Latex formula$ from the point $Latex formula$ ;

b) Find the points of intersection of this chord with the parabola.

c) Show that the chord of contact is a focal chord

### Solution 7

We firstly show that the point is external to the parabola. Upon substituting the point we obtain, LHS $Latex formula$, RHS $Latex formula$. Obviously LHS $Latex formula$ RHS, and hence we have that as $Latex formula$, and it follows that the point is external to the parabola.

a) The equation of the chord of contact from the external point $Latex formula$ on the parabola $Latex formula$ is given by

$Latex formula$

In this case , and hence . Thus, the equation of the chord of contact is,

In this case $Latex formula$, and hence $Latex formula$. Thus, the
equation of the chord of contact is:

$Latex formula$

i.e.

$Latex formula$

b) To find the points of intersection we must solve simultaneously the equations of the parabola and the chord itself. Doing so gives,

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

Hence we have the points of intersection being,

$Latex formula$

c) To show that the chord of contact is indeed a focal chord, we must substitute the coordinates of the foci into the equation for the chord of contact. Alternatively, we may find the parameters of the points of intersection of the chord of contact with the parabola and show that these parameters indeed have a product of $Latex formula$.

Now, the focus has coordinates $Latex formula$, and as $Latex formula$, then we have that the focus has coordinates $Latex formula$.

Substituting this into the equation of the chord of contact gives,

LHS $Latex formula$ RHS.

Hence $Latex formula$ lies on the chord of contact, and consequently the chord of contact is a focal chord.

## Important Geometrical Properties of the Parabola

In this section we consider two important geometrical properties of the parabola. These properties are of immense importance and are required knowledge by the HSC Mathematics Extension 1 syllabus. We shall prove the properties by considering them as examples.

### Example 8

(Focus-directrix property) Prove that the tangents drawn from the ends of a focal chord always intersect at right angles on the directrix.

### Solution 8

Consider the parabola $Latex formula$, where $Latex formula$, and the points $Latex formula$ and $Latex formula$ that lie on it. By definition, it follows that the directrix has equation $Latex formula$ and the focus has coordinates $Latex formula$.

Also, suppose that the tangents at $Latex formula$ and $Latex formula$ intersect at the point $Latex formula$.Firstly, we have that the chord $Latex formula$ has equation:

$Latex formula$

Since this chord passes through the point $Latex formula$, it then follows that

$Latex formula$

$Latex formula$

Now, the tangents at $Latex formula$and $Latex formula$ have equations

$Latex formula$

$Latex formula$

Respectively.

Solving these two equations simultaneously gives the point $Latex formula$. Thus $Latex formula$ gives,

$Latex formula$

$Latex formula$

$Latex formula$

Now, substituting this into $Latex formula$ gives,

$Latex formula$

$Latex formula$

Hence the point $Latex formula$ is given by

$Latex formula$

Now, from $Latex formula$ and hence we have that the point $Latex formula$ is given by

$Latex formula$

Thus the directrices intersect on the line $Latex formula$ which is the directrix of the parabola.

Now, to show that the tangents intersect at right angles, we consider the gradients of the tangents. The gradients of the tangents at $Latex formula$ and $Latex formula$ are $Latex formula$ and $Latex formula$ respectively, upon comparison to the standard $Latex formula$ form. However, it is known that $Latex formula$ by # and hence the gradients of the tangents have a product of $Latex formula$. Thus the tangents intersect at right angles.

Hence if a chord is a focal chord then the tangents drawn from the endpoints of this focal chord intersect at right angles on the directrix.

### Example 9

(Reflection Property) Prove that a ray of light parallel to the axis entering a parabolic mirror passes through the focus after reflection from the mirror.

### Solution 9

Consider the parabola $Latex formula$ and the point $Latex formula$ lying upon it. Suppose that the tangent at the point $Latex formula$ intersects the axis of the parabola with at the point $Latex formula$. Consider the diagram below.

For the light ray to pass through the focus, it must be reflected by the tangent at the point $Latex formula$ towards the focus. However, this will occur if and only if the angle of incidence is equal to the angle of reflection with respect to the tangent as the normal surface (This is the law of reflection from physics). So we are required to show that $Latex formula$.

Now, the axis of the parabola (The $Latex formula$-axis) is parallel to the incoming light ray. It hence follows that $Latex formula$, as these two angles form corresponding angles with the axis being parallel to the incoming light ray.

Thus, by proving that $Latex formula$ we can effectively show that $Latex formula$. Now, it easier to show that two lengths are equal then to show that two angles are equal in analytic geometry. Thus, by showing that $Latex formula$ we effectively show that [/latex] ?STP=?SPT[/latex]. The reason behind this is that the angles effectively form the base angles of an isosceles triangle, if $Latex formula$.

Now, we firstly find $Latex formula$.

$Latex formula$

$Latex formula$

$Latex formula$

Now we need to find the length of $Latex formula$. Before doing so, we must find the coordinates of the point $Latex formula$.

The tangent at the point has equation $Latex formula$. $Latex formula$ is the point of intersection of the tangent at $Latex formula$ with the axis of the parabola. Letting $Latex formula$, gives

$Latex formula$

And hence the point $Latex formula$ is

$Latex formula$

Hence we have that $Latex formula$

Thus as $Latex formula$, it thus follows that $Latex formula$. But from the argument above $Latex formula$. Hence it follows that $Latex formula$.

That is, a ray of light parallel to the axis entering a parabolic mirror passes through the focus after reflection from the mirror.

A common mistake with this proof is that students often attempt to prove the result purely through algebra using gradients. This is not only tedious and time consuming, but is also many times more difficult than the proof presented in these notes. Students should note that at times a purely algebraic proof is not the best, and if possible geometry should be introduced to simplify the process. The HSC Mathematics Extension courses are about the interplay between geometry, algebra, arithmetic and calculus. Thus a good understanding of these courses and their subtopics, requires the ability to apply each and every one of these aspects when required.

The two properties already investigated are important, and may be directly examined. However to ensure a complete understanding of the techniques involved we shall investigate some more examples.

### Example 10

The points $Latex formula$, $Latex formula$ and $Latex formula$ are points on the parabola $Latex formula$. Find the relation between $Latex formula$ and $Latex formula$ if the chord $Latex formula$ is perpendicular to the normal at $Latex formula$.

### Solution 10

Now, we firstly find the gradient of the parabola at the point $Latex formula$.

$Latex formula$

$Latex formula$

At the point $Latex formula$, the gradient is

$Latex formula$

Now the gradient of the normal at $Latex formula$ is given by,

$Latex formula$

The gradient of the chord $Latex formula$ is given by,

$Latex formula$

So we have that the chord $Latex formula$ is perpendicular to the normal at $Latex formula$. Thus the product of the gradients must be equal to $Latex formula$.

$Latex formula$

i.e.

$Latex formula$

$Latex formula$

Hence the relationship between $Latex formula$ and $Latex formula$ is that the sum of $Latex formula$ and $Latex formula$ equals twice $Latex formula$.

### Example 11

The point $Latex formula$ lies on the parabola $Latex formula$. $Latex formula$ is the midpoint of the chord $Latex formula$. Find the point $Latex formula$ on the parabola where the tangent is parallel to the chord $Latex formula$, and also show that $Latex formula$ is equidistant from $Latex formula$ and the $Latex formula$-axis.

### Solution 11

Firstly, we draw a diagram to show the given information.We firstly find the coordinates of the midpoint $Latex formula$. Doing so gives,Now, we find the gradient of $Latex formula$ to then subsequently solve for the point $Latex formula$.

$Latex formula$

We now need to find the point $Latex formula$, which is defined to have a tangent parallel to $Latex formula$. It is known that $Latex formula$ represents the gradient of the tangent at the point $Latex formula$. Thus, we need to solve the equation

$Latex formula$

for $Latex formula$. We have that

$Latex formula$

Thus, our equation becomes

$Latex formula$

As $Latex formula$ lies on the parabola, we thus have that,

$Latex formula$

Thus the point $Latex formula$ is given by

$Latex formula$

Now, to show that $Latex formula$ is equidistant from the point $Latex formula$ and the $Latex formula$-axis, we firstly find the perpendicular distance of the point $Latex formula$ from the $Latex formula$-axis. This is very simple, as the distance is simply the $Latex formula$-value of the coordinate of the point $Latex formula$.

Hence the distance from $Latex formula$ the -axis to the point $Latex formula$ is $Latex formula$.

Now, the distance from the point $Latex formula$ to $Latex formula$ is given by

$Latex formula$

Hence we have that the distance $Latex formula$ is equal to the perpendicular distance of the point $Latex formula$ from the $Latex formula$-axis.

## Locus problems relating to the Parabola

The locus of a point $Latex formula$ on the Cartesian ( $Latex formula$$Latex formula$ plane) is the geometrical path traced out by that point. The result is a set of points which form an algebraic equation. In this section we aim to obtain appropriate loci, given certain conditions by eliminating the parameter as required, to obtain an equation in terms $Latex formula$ of $Latex formula$ and only (this is called the “Cartesian equation”). Although the process may sound complicated, it is in fact quite straightforward. Consider the examples below, which illustrate the methods involved in these questions.

### Example 12

$Latex formula$ is a variable point on the parabola $Latex formula$. Find the locus of the midpoint of the chord $Latex formula$ as $Latex formula$ varies on the parabola. ( $Latex formula$ is the origin)

### Solution 12

Consider the diagram below, which shows the information given.

We have that the midpoint $Latex formula$ which has variable Cartesian coordinates $Latex formula$, has parametric coordinates,

$Latex formula$

Now, equating Cartesian and Parametric coordinates gives,

$Latex formula$

Now, we must eliminate the value of $Latex formula$. So we have,

$Latex formula$

Substituting this into the expression for $Latex formula$ gives,

$Latex formula$

Thus, the locus of $Latex formula$ as $Latex formula$ moves is,

$Latex formula$

### Example 13

The diagram shows the graph of the parabola $Latex formula$. The tangent to the parabola at $Latex formula$, $Latex formula$, cuts the $Latex formula$-axis at $Latex formula$. The normal to the parabola at $Latex formula$ cuts the $Latex formula$-axis at $Latex formula$.

a) Derive the equation of the tangent to the parabola at the point $Latex formula$.

b) Show that $Latex formula$ has coordinates $Latex formula$.

c) Let $Latex formula$ be the midpoint of $Latex formula$. Find the Cartesian equation of the locus of $Latex formula$.

### Solution 13

a) We have that

$Latex formula$

$Latex formula$

At the point $Latex formula$,

$Latex formula$

Hence we have the equation of the tangent,

$Latex formula$

$Latex formula$

b) We now need to find the equation of the normal to find the coordinates of the point $Latex formula$. The gradient of the normal at $Latex formula$ is equal to the negative reciprocal of the gradient of the tangent at $Latex formula$. So we have that,

$Latex formula$

So, using the point-gradient formula,

$Latex formula$

At the point $Latex formula$ we have that $Latex formula$, since this is the intersection with the $Latex formula$-axis.

$Latex formula$

Hence the point $Latex formula$ is given by $Latex formula$

c) The point $Latex formula$ is given by the intersection of the tangent with the $Latex formula$-axis. So, letting $Latex formula$ gives,

$Latex formula$

Hence the point $Latex formula$ is given by $Latex formula$.

The point $Latex formula$ in Cartesian coordinates, which is the midpoint of $Latex formula$ is given in parametric form by

$Latex formula$

Equating coordinates gives,

$Latex formula$

Thus, substituting for $Latex formula$ gives,

$Latex formula$

Thus, the locus of the point $Latex formula$ is

$Latex formula$

### Example 14

$Latex formula$ and $Latex formula$ are two variable points on a parabola $Latex formula$.

a) If the variable chord $Latex formula$ is always parallel to the line $Latex formula$, show that $Latex formula$.

b) The normals at $Latex formula$ and $Latex formula$ meet at $Latex formula$. Prove that the locus of $Latex formula$ is a straight line.

### Solution 14

Firstly, we draw a diagram to illustrate the information given.

a) The gradient of the chord $Latex formula$, $Latex formula$ is given by,

$Latex formula$

Now, this gradient must b equal to $Latex formula$ as the chord $Latex formula$ is at all times parallel to the lie $Latex formula$.

$Latex formula$

b) From the derivation in the “tangents, normals and chords”, the equation of the normal at the point $Latex formula$ is given by,

$Latex formula$

And the normal at $Latex formula$ is given by,

$Latex formula$

$Latex formula$ gives,

$Latex formula$

$Latex formula$

$Latex formula$

Now, at this point, to find the $Latex formula$ coordinate, we may substitute the $Latex formula$-value into $Latex formula$. However, this is tedious and an easier method is

$Latex formula$

Which gives,

$Latex formula$

Hence the point $Latex formula$ is given by

$Latex formula$

Equating Cartesian and parametric coordinates gives,

$Latex formula$

Now, at this point we perfect the square for the expression for $Latex formula$

$Latex formula$

Now, recall that $Latex formula$. Doing so gives,

$Latex formula$

$Latex formula$

We now eliminate $Latex formula$ from the expressions. Compare this to elimination of a single parameter used in earlier examples. This gives,

$Latex formula$

$Latex formula$

which is of course a straight line.

The technique employed here is an important one and is a vital member of your collection of techniques.

### Example 15

$Latex formula$ and $Latex formula$ are two points with parameters $Latex formula$ and $Latex formula$ respectively on the parabola $Latex formula$. The chord $Latex formula$passes through the fixed point $Latex formula$.

a) Show that $Latex formula$ $Latex formula$.

b) Find the locus of $Latex formula$, the point of intersection of the tangents $Latex formula$ at and $Latex formula$.

### Solution 15

a) Recall that the equation of the chord $Latex formula$ on the chord $Latex formula$ with parameters $Latex formula$ and $Latex formula$ respectively is given by:

$Latex formula$

Now, substituting the point $Latex formula$ into the equation gives,

$Latex formula$

$Latex formula$

b) Recall that the tangent at $Latex formula$ with parameter $Latex formula$ on the parabola $Latex formula$ has equation,

$Latex formula$

Similarly, the tangent at $Latex formula$ with parameter $Latex formula$ has equation,

$Latex formula$

Now, $Latex formula$ gives,

$Latex formula$

$Latex formula$

Substituting back into $Latex formula$ gives,

$Latex formula$

Hence the point of intersection $Latex formula$ has coordinates,

$Latex formula$

Now, equating Cartesian and parametric coordinates gives,

$Latex formula$

Now, recall from (a) that $Latex formula$ $Latex formula$.

So we have that,

$Latex formula$

$Latex formula$

Now, since the value of $Latex formula$ is a constant value regardless of the value of $Latex formula$, it thus follows that the locus of the point $Latex formula$ is the line $Latex formula$, since the values of $Latex formula$ and $Latex formula$ may vary through the real numbers.

A note should be made about the very common mistake of trying to eliminate the parameters without using the information given. Many students will try to find expressions for $Latex formula$ and $Latex formula$ by solving simultaneously. Do not fall into this common pitfall. This method is cumbersome and frivolous, as it leads nowhere. Always use the extra information given in the question.