Mathematics (2 unit) – Locus

Mathematics (2 unit) – Locus

The algebraic equation describing the geometrical path traced out by a variable point $Latex formula$ is called the locus of the point $Latex formula$. In this topic we shall study the loci involving the parabola, circles and straight lines.

Typically, problems involving loci are stated as “Find the locus of the point $Latex formula$ that moves such that a certain condition is obeyed”. The question is simply asking for the algebraic equation (Cartesian equation) that describes the geometrical path followed by the point that obeys that condition.

Loci involving straight lines

Locus problems involving straight lines are relatively easy. However, care must be taken in interpreting the question correctly, as this may result in errors. Observe the below examples to illustrate obtaining loci involving straight lines.

Example 1

Find the locus of the set of points which move such that $Latex formula$ is always 3 units to the left of the $Latex formula$-axis.

Solution 1

We draw a diagram to illustrate the path followed by $Latex formula$.

Now, if the point $Latex formula$ is at all times $Latex formula$ units to the left of the $Latex formula$ -axis, then all the points $Latex formula$units to the left form the locus. That is, the locus is $Latex formula$.

Example 2

Find the locus of the set of points which move such that $Latex formula$ is equidistant from the axes.

Solution 2

We firstly draw a diagram.

Now, we have that $Latex formula$ since the point $Latex formula$ is at all times equidistant from the coordinate axes.

So, using the distance formula we obtain,

$Latex formula$
$Latex formula$
$Latex formula$

Upon squaring both sides.

Thus the locus of the point $Latex formula$ is $Latex formula$. (Of course this simplifies down to $Latex formula$ which is equal to$Latex formula$) )

Note: This technique of squaring both sides is particularly important within this topic. Be sure to understand the use of it, and any particular intricacies it may have. A particular difficulty posed to students is when there is a coefficient placed in front of the square root implying use of the index law $Latex formula$. Most students however, forget to raise the $Latex formula$ to the power of $Latex formula$.

Example 3

Find the locus of the point $Latex formula$ that is equidistant from the points $Latex formula$ and $Latex formula$.

Solution 3

Firstly, let the point $Latex formula$ be $Latex formula$ and the point $Latex formula$ be $Latex formula$. We draw a diagram to depict the situation pictorially;

The condition that must be satisfied is,

$Latex formula$

Now, using the distance formula to describe $Latex formula$ and $Latex formula$ gives;

$Latex formula$

Squaring both sides, gives,

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

i.e.

$Latex formula$

Hence the locus of the point $Latex formula$ is $Latex formula$

Note: Locus questions such as the one above can be particularly tricky, and if the distances between the point $Latex formula$ and $Latex formula$, $Latex formula$ and $Latex formula$ are not equidistant, then the locus will be a circle instead. We shall see an example of this later on throughout the notes.

Loci involving Circles

A circle is defined to be the set of all points $Latex formula$ such that $Latex formula$ is equidistant from a point, say $Latex formula$. The point $Latex formula$ is called the center of the circle and the distance between $Latex formula$ and $Latex formula$ is called the radius.

Suppose that$Latex formula$ has coordinates $Latex formula$ and the distance between A and P is r. We can formulate the general equation of a circle (this should already be known) by simply using the definition stated above.

So we have that, the distance of the point $Latex formula$ from the point A is at all times equal to r. That is,

Using the distance formula to represent PA gives;

Now, squaring both sides of the expression gives;

 $Latex formula$

which is the general Cartesian (algebraic) equation of the circle with centre $Latex formula$ and radius $Latex formula$. Now suppose that the centre of the circle is at the origin. That is $Latex formula$. Hence we have that the equation of the circle with centre $Latex formula$ is;

 $Latex formula$

We shall now consider some examples to illustrate use of the formulae.

Example 4

Find the equation of the circle centre $Latex formula$, radius $Latex formula$.

Solution 4

We know that the equation of the circle, centre $Latex formula$, radius $Latex formula$ is given by $Latex formula$. Applying this formula with the following alterations, $Latex formula$ and $Latex formula$ gives the algebraic equation of the required circle as;$Latex formula$$Latex formula$

Note: when finding the equation of a circle, the best form to leave the equation in is the form $Latex formula$. There is no need to expand the squares. The reasoning behind this is that the centre and the radius may be readily found by observing the equation left in the desired form. Any expanded equation will require extra work to find the centre and radius.

Example 5

Find the centre and radius of the circle . $Latex formula$

Solution 5

By comparison with the general equation, it can be seen that the centre is the point $Latex formula$ and the radius $Latex formula$.

Example 6

Find the centre and radius of the circle $Latex formula$.

Solution 6

The above equation is not in general form, but is the expanded form of an equation of a circle. So to obtain the centre and radius, we must reverse the expansion, and complete the square, to obtain the equation in general form. Recall that $Latex formula$

$Latex formula$

Now, to complete the square for the $Latex formula$ bracket, we must simply add the square of half of the coefficient of $Latex formula$ to both sides (we must add to both sides to balance the equation). That is half of $Latex formula$, squared, which is simply $Latex formula$. To complete the square in the $Latex formula$ bracket, we must add the square of half of the coefficient of $Latex formula$ which in this case is the square of half of $Latex formula$ which is $Latex formula$ to both sides. Adding these values gives,

$Latex formula$

That is,

$Latex formula$

Hence by comparison to the general form, we have that the centre and radius of the circle is $Latex formula$ and $Latex formula$ respectively.

Example 7

Find the locus of the point $Latex formula$ which is always $Latex formula$ units from the origin $Latex formula$.

Solution 7

To find this locus, we may use previously developed formulae, but we shall instead use first principles here. The restriction is that,$Latex formula$Hence, using the distance formula to represent $Latex formula$ gives,

$Latex formula$

Upon squaring both sides we obtain,

$Latex formula$

Geometrically, this locus describes a circle centre $Latex formula$ with radius $Latex formula$.

Example 8

Find the locus of the point $Latex formula$ that moves such that the distance of $Latex formula$ from $Latex formula$ is twice the distance from $Latex formula$. Describe this locus geometrically.

Solution 8

With this locus, we are looking at the set of points that follow the rule $Latex formula$. Using the distance formula gives,$Latex formula$Squaring both sides gives;

$Latex formula$

$Latex formula$

$Latex formula$

Expanding gives;

$Latex formula$

Simplifying and collecting like terms gives;

$Latex formula$

Now, dividing through by $Latex formula$ gives;

$Latex formula$

Completing the square gives;

$Latex formula$

By comparison with the general form of the circle, it can be seen that the locus is geometrically a circle with centre $Latex formula$ and radius $Latex formula$.

Example 9

$Latex formula$ and $Latex formula$ are the points $Latex formula$ and $Latex formula$ respectively. Find the equation of the locus of the point $Latex formula$ which moves such that $Latex formula$.

Solution 9

To find this locus, we simply use the distance formula to find the value of the LHS and then simply rearrange and simplify to obtain the equation. Using the distance formula gives,

$Latex formula$

$Latex formula$

$Latex formula$

Now this is equal to 16. Equating expressions gives,

$Latex formula$

$Latex formula$ which is the locus of the point $Latex formula$

Loci involving the Parabola

The locus of the parabola is of the simplest non-degenerate conic sections to study. The definition of a parabola is the locus of a point $Latex formula$ which moves such that its distance from a given point (called the focus) is equal to its perpendicular distance from a line (called the directrix). We can derive the most commonly encountered equation of the parabola given this definition.

Consider the diagram above. This diagram depicts the most basic form of the parabola, and it is this case which we shall derive. However, other parabolas are derived from first principles using very similar techniques.

Let $Latex formula$ be the focus of the parabola and let the vertex of the parabola occur at the origin. As the vertex of the parabola occurs at the origin, then it follows that the directrix must be equal to $Latex formula$, since we have that $Latex formula$ must be equal to the perpendicular distance of the vertex from directrix. $Latex formula$ is the foot of the perpendicular from $Latex formula$ to the directrix. So we have by the definition of the locus of a parabola that,

$Latex formula$

Now, using the perpendicular distance formulae and some very simple coordinate geometry we can see that,

$Latex formula$

and that

$Latex formula$

&

Now, it is obvious that $Latex formula$, since we have that $Latex formula$ and obviously $Latex formula$ since it is equal to the $Latex formula$ -coordinate of the point $Latex formula$. Also, $Latex formula$ since we have that $Latex formula$ lies on the line $Latex formula$.

Hence we have that,

$Latex formula$

Squaring both sides gives,

$Latex formula$

That is,

$Latex formula$

Now, collecting like terms and cancelling from both sides gives,

 $Latex formula$

which is the simplest form of the parabola. You need not use the first principles method to derive the equation of a parabola, unless of course it is stated clearly in the question.

Note:
The absolute value of $Latex formula$ is called the focal length of the parabola, as it is the distance between the vertex and the focus. It is also the distance between the vertex and directrix. From this we can deduce that the vertex is the midpoint of the perpendicular from the directrix to the focus. You need not use the first principles method to derive the equation of a parabola, unless of course it is stated clearly in the question.

We shall now investigate the general equation of the four mainly encountered parabolas. Below are diagrams with the corresponding equation and position of the vertex. In each case the vertex has coordinates $Latex formula$, the focus is designated $Latex formula$ and the directrix is indicated by a dotted line. It should be noted that the four formulae introduced here for the four differing types of parabolas must be committed to memory (verbatim).

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

To solve problems involving the focus and directrix of a certain parabola, simply sketch the required parabola (a good diagram is a prerequisite) and then simply rearrange the Cartesian equation to obtain one of the four general forms. After this simply read off the focal length and then find the focus and the directrix by simply regarding the concavity of the curve.

We shall now consider some examples requiring use of the formulas above and the first principles definition of the parabola.

Example 10

Using first principles, find the equation of a parabola with focus $Latex formula$ and vertex $Latex formula$.

Solution 10

We firstly draw a diagram, labelling the points given.It can be seen that the curve looks like so since the concave side of the curve contains the focus. Now, since there is a 2 unit difference between the focus and the vertex, then the focal length is 2 and hence the directrix lies 2 units to the right of the vertex and has equation $Latex formula$. Now, we simply use the definition to obtain the equation of the parabola.$Latex formula$$Latex formula$

Now, by the definition of the parabola $Latex formula$. Hence we have that,

$Latex formula$

Squaring both sides gives,

$Latex formula$

Expanding the terms in $Latex formula$ gives,

$Latex formula$

Cancelling terms gives,

$Latex formula$

$Latex formula$

Which is the equation of the parabola.

Example 11

A parabola has equation $Latex formula$

1. Find the vertex and the focus.
2. What is the equation of the directrix?

Solution 11

1) So we simply perfect the square, and read off the vertex after comparing to one of the four general forms.

$Latex formula$

$Latex formula$

$Latex formula$

$Latex formula$

Hence the vertex of the parabola is $Latex formula$ after comparison to the first form above and that the focal length is $Latex formula$.

Now, we sketch a diagram to see the position of the focus.

Since the focal length is $Latex formula$ (by rearranging the formula into a standard form) we can see that the focus will be $Latex formula$ units above the vertex and hence has coordinates $Latex formula$.

2) Now, since the directrix is a focal length below the vertex in this case, then the directrix has equation $Latex formula$.

Example 12

$Latex formula$ and $Latex formula$ are the points $Latex formula$ and $Latex formula$ respectively. Show that AB is a focal chord of $Latex formula$.

Solution 12

Firstly, we can see that the vertex of the parabola is the origin. Now, after rearranging the equation of the parabola, we see that it is equivalent to $Latex formula$

Hence we have that $Latex formula$ and that the focal length is $Latex formula$.Hence we have that the focus has coordinates $Latex formula$. (Check this by drawing a diagram!) Now, to show that the line $Latex formula$ is a focal chord, we simply show that $Latex formula$ passes through the focus.

Now, to find the equation of $Latex formula$ we simply find the gradient, then use the point gradient formula.

$Latex formula$

Now, using the point gradient formula gives,

$Latex formula$

Now, substituting the point $Latex formula$ gives,

$Latex formula$

$Latex formula$

Hence we have that the line $Latex formula$ is a focal chord to the parabola $Latex formula$.

Example 13

Find the focus, directrix and vertex of the parabola $Latex formula$.

Solution 13

Firstly, we notice that this equation has a square in the $Latex formula$ and hence we shall be completing the square in the $Latex formula$ component of the equation. Doing so gives $Latex formula$

$Latex formula$

That is,

$Latex formula$

Now, it is obvious that $Latex formula$ and that the coordinates of the vertex are $Latex formula$. We now draw a simple sketch of the parabola to assist with the obtaining of the focus and the directrix.

Now, since the focal length is $Latex formula$ we thus have that the coordinates of the focus are $Latex formula$ and that the equation of the directrix is $Latex formula$.