Mathematics (2 unit) – Locus
The algebraic equation describing the geometrical path traced out by a variable point is called the locus of the point . 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 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 is always 3 units to the left of the -axis.
Solution 1
We draw a diagram to illustrate the path followed by .
Now, if the point is at all times units to the left of the -axis, then all the points units to the left form the locus. That is, the locus is .
Example 2
Find the locus of the set of points which move such that is equidistant from the axes.
Solution 2
We firstly draw a diagram.
Now, we have that since the point is at all times equidistant from the coordinate axes.
So, using the distance formula we obtain,
Upon squaring both sides.
Thus the locus of the point is . (Of course this simplifies down to which is equal to) )
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 . Most students however, forget to raise the to the power of .
Example 3
Find the locus of the point that is equidistant from the points and .
Solution 3
Firstly, let the point be and the point be . We draw a diagram to depict the situation pictorially;
The condition that must be satisfied is,
Now, using the distance formula to describe and gives;
Squaring both sides, gives,
i.e.
Hence the locus of the point is
Note: Locus questions such as the one above can be particularly tricky, and if the distances between the point and , and 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 such that is equidistant from a point, say . The point is called the center of the circle and the distance between and is called the radius.
Suppose that has coordinates 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 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;
which is the general Cartesian (algebraic) equation of the circle with centre and radius . Now suppose that the centre of the circle is at the origin. That is . Hence we have that the equation of the circle with centre is;
We shall now consider some examples to illustrate use of the formulae.
Example 4Find the equation of the circle centre , radius . Solution 4We know that the equation of the circle, centre , radius is given by . Applying this formula with the following alterations, and gives the algebraic equation of the required circle as; |
Note: when finding the equation of a circle, the best form to leave the equation in is the form . 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 5Find the centre and radius of the circle . Solution 5By comparison with the general equation, it can be seen that the centre is the point and the radius . |
Example 6Find the centre and radius of the circle . Solution 6The 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
Now, to complete the square for the bracket, we must simply add the square of half of the coefficient of to both sides (we must add to both sides to balance the equation). That is half of , squared, which is simply . To complete the square in the bracket, we must add the square of half of the coefficient of which in this case is the square of half of which is to both sides. Adding these values gives,
That is,
Hence by comparison to the general form, we have that the centre and radius of the circle is and respectively. |
Example 7Find the locus of the point which is always units from the origin . Solution 7To find this locus, we may use previously developed formulae, but we shall instead use first principles here. The restriction is that,Hence, using the distance formula to represent gives,
Upon squaring both sides we obtain,
Geometrically, this locus describes a circle centre with radius . |
Example 8Find the locus of the point that moves such that the distance of from is twice the distance from . Describe this locus geometrically. Solution 8With this locus, we are looking at the set of points that follow the rule . Using the distance formula gives,Squaring both sides gives;
Expanding gives;
Simplifying and collecting like terms gives;
Now, dividing through by gives;
Completing the square gives;
By comparison with the general form of the circle, it can be seen that the locus is geometrically a circle with centre and radius . |
Example 9and are the points and respectively. Find the equation of the locus of the point which moves such that . Solution 9To 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,
Now this is equal to 16. Equating expressions gives,
which is the locus of the point |
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 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 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 , since we have that must be equal to the perpendicular distance of the vertex from directrix. is the foot of the perpendicular from to the directrix. So we have by the definition of the locus of a parabola that,
Now, using the perpendicular distance formulae and some very simple coordinate geometry we can see that,
and that
&
Now, it is obvious that , since we have that and obviously since it is equal to the -coordinate of the point . Also, since we have that lies on the line .
Hence we have that,
Squaring both sides gives,
That is,
Now, collecting like terms and cancelling from both sides gives,
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 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 , the focus is designated 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).
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 11A parabola has equation
Solution 111) So we simply perfect the square, and read off the vertex after comparing to one of the four general forms.
Hence the vertex of the parabola is after comparison to the first form above and that the focal length is . Now, we sketch a diagram to see the position of the focus. Since the focal length is (by rearranging the formula into a standard form) we can see that the focus will be units above the vertex and hence has coordinates . 2) Now, since the directrix is a focal length below the vertex in this case, then the directrix has equation . |
Example 12and are the points and respectively. Show that AB is a focal chord of . Solution 12Firstly, 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 Hence we have that and that the focal length is .Hence we have that the focus has coordinates . (Check this by drawing a diagram!) Now, to show that the line is a focal chord, we simply show that passes through the focus. Now, to find the equation of we simply find the gradient, then use the point gradient formula.
Now, using the point gradient formula gives,
Now, substituting the point gives,
Hence we have that the line is a focal chord to the parabola . |
Example 13Find the focus, directrix and vertex of the parabola . Solution 13Firstly, we notice that this equation has a square in the and hence we shall be completing the square in the component of the equation. Doing so gives
That is,
Now, it is obvious that and that the coordinates of the vertex are . 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 we thus have that the coordinates of the focus are and that the equation of the directrix is . |