2008 Hsc Mathematics Extension 1 Sample Answers

Parametric Equations

Parametric Representation

Contents

• 1 Parametric Representation
• 2 Converting From Parametric Form to Cartesian form and vice versa
  • 2.1 Example 1
  • 2.2 Solution 1
  • 2.3 Example 2
  • 2.4 Solution 2
  • 2.5 Example 3
  • 2.6 Solution 3
• 3 Tangents, Normals and Chords
  • 3.1 Tangents
  • 3.2 Normals
  • 3.3 Chords
  • 3.4 Chord of Contact
  • 3.5 Example 4
  • 3.6 Solution 4
  • 3.7 Example 5
  • 3.8 Solution 5
  • 3.9 Example 6
  • 3.10 Solution 6
  • 3.11 Example 7
  • 3.12 Solution 7
• 4 Important Geometrical Properties of the Parabola
  • 4.1 Example 8
  • 4.2 Solution 8
  • 4.3 Example 9
  • 4.4 Solution 9
  • 4.5 Example 10
  • 4.6 Solution 10
  • 4.7 Example 11
  • 4.8 Solution 11
• 5 Locus problems relating to the Parabola
  • 5.1 Example 12
  • 5.2 Solution 12
  • 5.3 Example 13
  • 5.4 Solution 13
  • 5.5 Example 14
  • 5.6 Solution 14
  • 5.7 Example 15
  • 5.8 Solution 15

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

These equations are called parametric equations , and  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 in terms of two parametric equations. These equations are , . By the simple elimination of 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 are given by:

It should be noted that the sign of indicates whether or not the parabola is concave up or down. If  then it follows that the parabola is concave up, and if  then it follows that the parabola is concave down. Also, as a side note , the absolute value of the value of , 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 or the parabola with equation  and then substitute into the equations,

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

We would simply perform the following algebraic manipulations,

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 .

Solution 1

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

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

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 , .

Solution 2

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

Hence, upon substitution we obtain,

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 .

Solution 3

Firstly we rearrange the equation to obtain the value of .

Hence the value of  is .

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

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

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  lying on the parabola . 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  is dubbed the Parametric form of the tangent.

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

The parametric form of the equation of the tangent at the point is,

Normals

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

The parametric form of the equation of the normal at the point  is,

Chords

The equation of the chord between the points  and  on the parabola  is

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

That is,

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

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 . This fact becomes important later on.

Chord of Contact

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

To show that a point  is external to the parabola, we simply substitute the point into the equation , and if  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 to the parabola  has equation . 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  on the parabola  is,

We shall now consider some examples.

Example 4

Find the equations of the tangents to the parabola  at the points  and. 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, . Hence we have that the equation of the tangent at the point where coordinates are given by:

When  we have the equation of the tangent being,

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

Hence the two tangents intersect at the point

Example 5

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

a) The coordinates of and

b) The equations of the normals at  and

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,

Hence the coordinates of  and  are,

 and

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 :

Also, the normal at  has equation

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

gives,

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

Example 6

Find the equation of the chord joining  and  on and :

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  and .

When , . Hence the point is .

When Hence the point is:

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

Now, using the point gradient formula,

Hence the equation of  is

b) Now, the formulae derived is

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

That is,

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

Now, substituting into the equation of the chord gives:

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 . That is, . Recall that this was found to be a necessary and sufficient condition for the chord  to be a focal chord.

Example 7

Show that the point  is external to the parabola . Hence,

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

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 , RHS . Obviously LHS RHS, and hence we have that as , and it follows that the point is external to the parabola.

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

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

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

i.e.

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

Hence we have the points of intersection being,

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 .

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

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

LHS  RHS.

Hence  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 , where , and the points  and  that lie on it. By definition, it follows that the directrix has equation  and the focus has coordinates .

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

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

Now, the tangents at and have equations

Respectively.

Solving these two equations simultaneously gives the point . Thus  gives,

Now, substituting this into  gives,

Hence the point  is given by

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

Thus the directrices intersect on the line  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  and  are  and  respectively, upon comparison to the standard  form. However, it is known that by # and hence the gradients of the tangents have a product of . 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  and the point  lying upon it. Suppose that the tangent at the point  intersects the axis of the parabola with at the point . Consider the diagram below.

For the light ray to pass through the focus, it must be reflected by the tangent at the point  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 .

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

Thus, by proving that  we can effectively show that . 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  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 .

Now, we firstly find .

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

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

And hence the point  is

Hence we have that

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

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

Solution 10

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

At the point , the gradient is

Now the gradient of the normal at  is given by,

The gradient of the chord  is given by,

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

i.e.

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

Example 11

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

Solution 11

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

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

for . We have that

Thus, our equation becomes

As  lies on the parabola, we thus have that,

Thus the point  is given by

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

Hence the distance from  the -axis to the point is.

Now, the distance from the point  to  is given by

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

Locus problems relating to the Parabola

The locus of a point  on the Cartesian ( –  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 of  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

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

Solution 12

Consider the diagram below, which shows the information given.

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

Now, equating Cartesian and Parametric coordinates gives,

Now, we must eliminate the value of . So we have,

Substituting this into the expression for  gives,

Thus, the locus of  as  moves is,

Example 13

Thediagram shows the graph of the parabola . The tangent to the parabola at , , cuts the -axis at . The normal to the parabola at  cuts the -axis at .

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

b) Show that  has coordinates .

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

Solution 13

a) We have that

At the point ,

Hence we have the equation of the tangent,

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

So, using the point-gradient formula,

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

Hence the point  is given by

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

Hence the point  is given by .

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

Equating coordinates gives,

Thus, substituting for  gives,

Thus, the locus of the point  is

Example 14

and  are two variable points on a parabola .

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

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

Solution 14

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

a) The gradient of the chord ,  is given by,

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

b) From the derivation in the "tangents, normals and chords", the equation of the normal at the point  is given by,

And the normal at  is given by,

gives,

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

Which gives,

Hence the point  is given by

Equating Cartesian and parametric coordinates gives,

Now, at this point we perfect the square for the expression for

Now, recall that . Doing so gives,

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

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

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

a) Show that .

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

Solution 15

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

Now, substituting the point  into the equation gives,

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

Similarly, the tangent at  with parameter  has equation,

Now,  gives,

Substituting back into  gives,

Hence the point of intersection  has coordinates,

Now, equating Cartesian and parametric coordinates gives,

Now, recall from (a) that .

So we have that,

Now, since the value of  is a constant value regardless of the value of , it thus follows that the locus of the point  is the line , since the values of  and  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  and  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.

2008 Hsc Mathematics Extension 1 Sample Answers

Source: https://dc.edu.au/mathematics-extension-1-parametric-equations/

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