Co-ordinate geometry in the (x, y) plane

Mathematics 2210 · CCEA A Level · 7 min read · Updated: 31 Aug 2026

Co-ordinate Geometry in the \((x, y)\) Plane: Parametric Equations

Welcome to A2 Coordinate Geometry! In AS Level, you mastered equations where \(y\) is written directly in terms of \(x\), such as \(y = mx + c\) or \((x - a)^2 + (y - b)^2 = r^2\). These are called Cartesian equations. In A2, we take our coordinate skills to the next level by introducing a powerful new way to describe curves: Parametric Equations.

Don't worry if this idea feels a bit unfamiliar at first. By breaking each method down into clear, step-by-step recipes, you will see how intuitive and enjoyable this topic can be!

1. What are Parametric Equations?

Instead of linking \(x\) and \(y\) directly in one single formula, we link both \(x\) and \(y\) to a third, independent variable called a parameter. The parameter is usually written as \(t\) (often representing time) or \(\theta\) (representing an angle).

Real-World Analogy: Imagine a drone flying through the air. A GPS tracker records its horizontal position \(x\) at time \(t\), and its vertical position \(y\) at time \(t\).
- \(x = f(t)\)
- \(y = g(t)\)
Together, these two separate equations describe the drone's entire flight path in the 2D plane!

Finding Points on a Parametric Curve

To find the coordinates of a point on the curve, simply plug the given value of the parameter into both equations.

Example: A curve has parametric equations \(x = 2t + 1\) and \(y = t^2 - 3\). Find the coordinates of the point where \(t = 3\).
- For \(x\): \(x = 2(3) + 1 = 7\)
- For \(y\): \(y = (3)^2 - 3 = 6\)
The coordinates of the point are \((7, 6)\).

Key Takeaway: Each value of the parameter gives you exactly one point \((x, y)\) on the curve.

2. Converting from Parametric to Cartesian Form

To convert parametric equations into Cartesian form (an equation with only \(x\) and \(y\)), our goal is to eliminate the parameter. There are two primary methods you need to know.

Method A: Rearrangement and Substitution (Algebraic)

Use this method when the equations contain standard algebraic terms (like powers or polynomials).

Step 1: Rearrange the simpler equation to make the parameter \(t\) the subject.
Step 2: Substitute this expression for \(t\) into the other equation.
Step 3: Simplify your final equation into the required format (e.g., \(y = f(x)\) or \(f(x, y) = 0\)).

Worked Example: Find the Cartesian equation of the curve given by \(x = \frac{1}{t - 2}\) and \(y = 3t + 1\) for \(t \neq 2\).
- Step 1: Rearrange \(x = \frac{1}{t - 2}\) for \(t\):
\(x(t - 2) = 1 \implies t - 2 = \frac{1}{x} \implies t = \frac{1}{x} + 2\)
- Step 2: Substitute \(t = \frac{1}{x} + 2\) into the equation for \(y\):
\(y = 3\left(\frac{1}{x} + 2\right) + 1\)
- Step 3: Expand and simplify:
\(y = \frac{3}{x} + 6 + 1 \implies y = \frac{3}{x} + 7\)

Method B: Using Trigonometric Identities

When the parameter is an angle \(\theta\) or \(t\) inside trigonometric functions, rearranging directly to find the angle can create awkward inverse trig functions. Instead, we rearrange for the trig ratios and use standard trigonometric identities to eliminate the parameter.

Essential Identities to Keep Handy:
- \(\cos^2 \theta + \sin^2 \theta = 1\)
- \(1 + \tan^2 \theta = \sec^2 \theta \implies \sec^2 \theta - \tan^2 \theta = 1\)
- Double angle identities, such as \(\cos 2\theta = 1 - 2\sin^2 \theta\) or \(\cos 2\theta = 2\cos^2 \theta - 1\)

Worked Example: A curve is defined by \(x = 3\cos \theta\) and \(y = 2\sin \theta\). Find its Cartesian equation.
- Step 1: Make \(\cos \theta\) and \(\sin \theta\) the subjects:
\(\cos \theta = \frac{x}{3}\) and \(\sin \theta = \frac{y}{2}\)
- Step 2: Use the identity \(\cos^2 \theta + \sin^2 \theta = 1\):
\(\left(\frac{x}{3}\right)^2 + \left(\frac{y}{2}\right)^2 = 1\)
- Step 3: Simplify:
\(\frac{x^2}{9} + \frac{y^2}{4} = 1\)
(Did you know? This is the standard Cartesian equation of an ellipse!)

Common Mistake to Avoid: Don't forget to square both the numerator and the denominator when squaring fractions like \(\left(\frac{x}{3}\right)^2 = \frac{x^2}{9}\).

Key Takeaway: To eliminate algebraic parameters, substitute \(t\). To eliminate trigonometric parameters, isolate the trig functions and use an identity like \(\sin^2 \theta + \cos^2 \theta = 1\).

3. Points of Intersection with Coordinate Axes and Lines

To find where a parametric curve crosses axes or lines, set up an equation using the coordinate condition first to find the value of the parameter \(t\), and then determine the full coordinates.

Intersections with the Axes:

- \(y\)-axis intersection: Set \(x = 0\), solve for \(t\), then calculate \(y\).
- \(x\)-axis intersection: Set \(y = 0\), solve for \(t\), then calculate \(x\).

Intersection with a Straight Line:

If you are finding the intersection of a curve \(x = f(t)\), \(y = g(t)\) with a line \(y = mx + c\):
- Substitute \(f(t)\) for \(x\) and \(g(t)\) for \(y\) directly into the line equation.
- Solve the resulting equation for \(t\).
- Substitute the value(s) of \(t\) back into \(x = f(t)\) and \(y = g(t)\) to find the coordinate pairs.

Quick Review: Always check if an exam question asks for "the value of the parameter \(t\)" or "the coordinates of the point \((x, y)\)". Many students lose easy marks by stopping once they find \(t\)!

4. Differentiation of Parametric Curves

To find the gradient of a curve given parametrically, we use the Chain Rule from calculus.

The Parametric Gradient Formula

Because \(y\) and \(x\) are both functions of \(t\):
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}\) (provided \(\frac{\mathrm{d}x}{\mathrm{d}t} \neq 0\))

Memory Tip: Think of it as "dividing fractions" where the \(\mathrm{d}t\) terms cancel out: \(\frac{\mathrm{d}y}{\mathrm{d}t} \div \frac{\mathrm{d}x}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}t} \times \frac{\mathrm{d}t}{\mathrm{d}x} = \frac{\mathrm{d}y}{\mathrm{d}x}\).

Step-by-Step Method for Finding the Gradient:

1. Differentiate \(x\) with respect to \(t\) to find \(\frac{\mathrm{d}x}{\mathrm{d}t}\).
2. Differentiate \(y\) with respect to \(t\) to find \(\frac{\mathrm{d}y}{\mathrm{d}t}\).
3. Divide \(\frac{\mathrm{d}y}{\mathrm{d}t}\) by \(\frac{\mathrm{d}x}{\mathrm{d}t}\) to get \(\frac{\mathrm{d}y}{\mathrm{d}x}\) in terms of \(t\).
4. Substitute the specific value of \(t\) to get the numerical gradient.

Stationary Points on Parametric Curves

A stationary point occurs when the gradient is zero:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = 0 \implies \frac{\mathrm{d}y}{\mathrm{d}t} = 0\) (provided \(\frac{\mathrm{d}x}{\mathrm{d}t} \neq 0\))

5. Tangents and Normals to Parametric Curves

Once you have the gradient \(\frac{\mathrm{d}y}{\mathrm{d}x}\), finding the equations of tangents and normals follows the exact same straight-line geometry you learned at AS Level!

Key Formulae:

- Equation of a Straight Line: \(y - y_1 = m(x - x_1)\)
- Gradient of Tangent (\(m_T\)): \(m_T = \frac{\mathrm{d}y}{\mathrm{d}x}\)
- Gradient of Normal (\(m_N\)): \(m_N = -\frac{1}{m_T}\) (perpendicular line rule: \(m_T \times m_N = -1\))

Complete Worked Example:

A curve has parametric equations \(x = t^2 + 2\), \(y = 4t - 1\).
Find the equation of the tangent and the normal to the curve at the point where \(t = 2\).

Step 1: Find the coordinates of the point \((x_1, y_1)\) when \(t = 2\)
\(x_1 = (2)^2 + 2 = 6\)
\(y_1 = 4(2) - 1 = 7\)
Point is \((6, 7)\).

Step 2: Differentiate to find \(\frac{\mathrm{d}y}{\mathrm{d}x}\)
\(\frac{\mathrm{d}x}{\mathrm{d}t} = 2t\)
\(\frac{\mathrm{d}y}{\mathrm{d}t} = 4\)
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}} = \frac{4}{2t} = \frac{2}{t}\)

Step 3: Calculate the gradient at \(t = 2\)
Gradient of tangent \(m_T = \frac{2}{2} = 1\)
Gradient of normal \(m_N = -\frac{1}{1} = -1\)

Step 4: Find the equation of the Tangent
Using \(y - y_1 = m_T(x - x_1)\):
\(y - 7 = 1(x - 6)\)
\(y - 7 = x - 6 \implies y = x + 1\)

Step 5: Find the equation of the Normal
Using \(y - y_1 = m_N(x - x_1)\):
\(y - 7 = -1(x - 6)\)
\(y - 7 = -x + 6 \implies y = -x + 13\) (or \(x + y - 13 = 0\))

Key Takeaway: Finding tangents and normals always needs three ingredients: \(x_1\), \(y_1\), and the gradient \(m\) at the specific value of \(t\).

6. Summary and Top Exam Tips

Quick Review Checklist:
- Parameter Conversion (Algebraic): Make \(t\) the subject of one equation and substitute into the other.
- Parameter Conversion (Trigonometric): Isolate the trig functions and use \(\sin^2 \theta + \cos^2 \theta = 1\) or \(\sec^2 \theta - \tan^2 \theta = 1\).
- Parametric Gradient: \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t}\).
- Tangents and Normals: Always find the point \((x_1, y_1)\) first before using \(y - y_1 = m(x - x_1)\).

Top Exam Tip: Keep your negative signs safe when working with perpendicular gradients (\(m_N = -\frac{1}{m_T}\)), and make sure you write the final line equation in the exact form requested in the question (e.g., \(y = mx + c\) or \(ax + by + c = 0\)).

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What are parametric equations and how do you find coordinates of points on a parametric curve?

Parametric equations express both \(x\) and \(y\) in terms of an independent third variable called a parameter (e.g., \(t\) or \(\theta\)). To find a point's coordinates, substitute the given parameter value into both \(x = f(t)\) and \(y = g(t)\).

How do you convert algebraic parametric equations into Cartesian form?

Rearrange one equation to make the parameter \(t\) the subject, substitute this expression into the other equation, and simplify to obtain a relationship containing only \(x\) and \(y\).

How do you eliminate a trigonometric parameter to find a Cartesian equation?

Isolate the trigonometric functions in terms of \(x\) and \(y\), then substitute them into a standard trigonometric identity, such as \(\cos^2 \theta + \sin^2 \theta = 1\) or \(\sec^2 \theta - \tan^2 \theta = 1\).

How do you find where a parametric curve intersects the coordinate axes or a given straight line?

Apply the geometric condition (e.g., \(x=0\) for the \(y\)-axis, \(y=0\) for the \(x\)-axis, or substitute \(x(t)\) and \(y(t)\) into the line equation) to solve for the parameter \(t\), then evaluate \(x\) and \(y\) to find the coordinates.

What is the formula for the gradient \(\frac{\mathrm{d}y}{\mathrm{d}x}\) of a parametric curve?

\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}\) (provided \(\frac{\mathrm{d}x}{\mathrm{d}t} \neq 0\)).

What condition identifies a stationary point on a parametric curve?

A stationary point occurs when \(\frac{\mathrm{d}y}{\mathrm{d}x} = 0\), which corresponds to \(\frac{\mathrm{d}y}{\mathrm{d}t} = 0\) (provided \(\frac{\mathrm{d}x}{\mathrm{d}t} \neq 0\)).

How do you find the equations of the tangent and normal to a parametric curve at a given parameter value?

Find \((x_1, y_1)\) by evaluating \(x(t)\) and \(y(t)\), compute \(m_T = \frac{\mathrm{d}y}{\mathrm{d}x}\) (with normal gradient \(m_N = -\frac{1}{m_T}\)), and substitute into \(y - y_1 = m(x - x_1)\).

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