💪 FP1.2 Coordinate Geometry: Finding Loci (The Rules of the Road)
Welcome to Coordinate Geometry! In Further Maths, we move beyond plotting points and lines to finding the equations of paths (or shapes) defined by specific geometric rules. This path is called a Locus.
This chapter is all about translating simple distance rules into a precise Cartesian equation (an equation in terms of \(x\) and \(y\)). Don't worry if it seems tricky at first; we only need two key skills: calculating the distance between points and the distance to a simple vertical or horizontal line.
1. Understanding the Concept of a Locus
What is a Locus?
The term Locus (plural: Loci) simply means the set of all points that satisfy a given geometrical condition. Think of it as mapping out a path based on a strict rule.
- Example 1: The locus of points equidistant from a single fixed point is a Circle.
- Example 2 (The Focus of this Chapter): The locus of points equidistant from a fixed point and a fixed line is a Parabola.
Key Takeaway: We are looking for an equation (in \(x\) and \(y\)) that describes every point \(P(x, y)\) that obeys the specified distance rule.
2. The Building Blocks: Simple Distance Calculations
To find a locus, you need to be able to accurately calculate distances. We assume our moving point is \(P(x, y)\).
2.1 Distance from a Point A to the Locus P
If you have a fixed point \(A(x_1, y_1)\) and the moving point \(P(x, y)\), the distance \(PA\) is found using the standard distance formula (which is just Pythagoras' theorem):
$$PA = \sqrt{(x - x_1)^2 + (y - y_1)^2}$$
2.2 Distance from a Straight Line L to the Locus P
The syllabus restricts us to very simple vertical or horizontal fixed lines. This is great news, as the distance calculation is straightforward!
The distance from a point \(P(x, y)\) to a line is always measured perpendicular to the line. Since distance must be positive, we use the absolute value (the modulus sign, \(| \dots |\)).
Case 1: Vertical Line (\(x = a\))
The distance from \(P(x, y)\) to the line \(x = a\) is simply the difference between the x-coordinates:
$$PL = |x - a|$$
Case 2: Horizontal Line (\(y = b\))
The distance from \(P(x, y)\) to the line \(y = b\) is the difference between the y-coordinates:
$$PL = |y - b|$$
Always remember that distance must be positive. If your point \(P\) is at \(x=1\) and the line is \(x=5\), the distance is \(|1 - 5| = |-4| = 4\). Using the modulus ensures we get a positive value, even if the point is on the "wrong" side of the line.
3. The Equidistant Locus: Point and Simple Line
The central task in this topic is finding the equation of the locus \(P(x, y)\) such that:
Distance from Point A = Distance from Line L
To eliminate the square root from the distance formula, we always square both sides immediately:
$$(\text{Distance } PA)^2 = (\text{Distance } PL)^2$$
3.1 Step-by-Step Example Derivation
Let's find the Cartesian equation of the locus of points \(P(x, y)\) that are equidistant from the point \(A(2, 3)\) and the line \(x = 4\).
- Identify the Components:
- Fixed Point \(A(x_1, y_1) = (2, 3)\)
- Fixed Line \(L\): \(x = 4\)
- Locus Point \(P(x, y)\)
- Set up the Distance Equations:
- Distance \(PA\): \(\sqrt{(x - 2)^2 + (y - 3)^2}\)
- Distance \(PL\): \(|x - 4|\)
- Equate and Square Both Sides:
$$(\sqrt{(x - 2)^2 + (y - 3)^2})^2 = (|x - 4|)^2$$
Note: Squaring the absolute value \(|x-4|\) gives \((x-4)^2\).
$$(x - 2)^2 + (y - 3)^2 = (x - 4)^2$$
- Expand and Simplify (This is where the magic happens):
Expand the bracket terms:
$$(x^2 - 4x + 4) + (y^2 - 6y + 9) = (x^2 - 8x + 16)$$
Notice the \(x^2\) terms on both sides cancel out. This is a tell-tale sign that the locus is a parabola (a quadratic relationship between the remaining variable, \(y^2\), and the linear variable, \(x\)).
$$x^2 - 4x + 4 + (y - 3)^2 = x^2 - 8x + 16$$
Now, isolate the term involving \(y\) and gather the \(x\) and constant terms:
$$(y - 3)^2 = -8x + 16 - (-4x + 4)$$
$$(y - 3)^2 = -8x + 16 + 4x - 4$$
$$(y - 3)^2 = -4x + 12$$
- Final Equation (Cartesian Form):
The equation of the locus is \((y - 3)^2 = -4(x - 3)\). This is the equation of a parabola.
Do not forget to square the distance from the line, \(PL\). If you forget the modulus signs when squaring, it usually doesn't matter, but conceptually, you must always be working with the positive distance first before squaring.
4. General Form of the Resulting Parabola
When you solve a locus problem involving a point (the focus) and a line (the directrix), the result will always be a parabola. The form of the final equation depends on whether the directrix line is horizontal or vertical.
If the directrix is \(x=a\) (vertical), the resulting parabola will be of the form:
$$\mathbf{(y - k)^2 = 4c(x - h)}$$
If the directrix is \(y=b\) (horizontal), the resulting parabola will be of the form:
$$\mathbf{(x - h)^2 = 4c(y - k)}$$
Did you know? In our example, the resulting equation \((y - 3)^2 = -4(x - 3)\) matches the standard form \((y - k)^2 = 4c(x - h)\). This means the vertex of the parabola is \((3, 3)\) and since \(4c = -4\) (so \(c = -1\)), the parabola opens to the left, away from the directrix \(x=4\). The geometry works out perfectly!
5. Summary and Key Takeaway
The goal is to find the Cartesian equation \(P(x, y)\) based on the rule:
1. Rule: Distance \(PA\) = Distance \(PL\).
2. Method: Square both sides immediately: \((PA)^2 = (PL)^2\).
3. Formulas:
- \((PA)^2 = (x - x_1)^2 + (y - y_1)^2\)
- \((PL)^2 = (x - a)^2\) (if line is \(x=a\))
- \((PL)^2 = (y - b)^2\) (if line is \(y=b\))
4. Result: Expand, simplify, and ensure you arrange the equation neatly, often cancelling the quadratic term for one variable to reveal the parabola.
Coordinate geometry is all about applying algebraic tools to geometric problems. By mastering the distance formula and the simple distance-to-line calculations, you can easily handle the locus problems in FP1!