# Coordinate Systems (FP1)

> Edexcel International A-Level Further Mathematics · FP1
> Source: https://www.owlsprep.com/study/edexcel-ial-further-math-fp1-coordinate-systems/

This guide covers all Edexcel IAL FP1 coordinate systems content: standard and parametric forms of parabolas and rectangular hyperbolas, focus-directrix properties of parabolas, and calculating tangents/normals using Cartesian differentiation.

**Prerequisites:** [P1 differentiation (power rule for fractional and negative exponents)](https://www.owlsprep.com/study/edexcel-ial-maths-p1-differentiation/); [Equations of straight lines ($y - y_1 = m(x - x_1)$)](https://www.owlsprep.com/study/edexcel-ial-maths-p1-straight-line-graphs/)

## Learning objectives

- Recall standard Cartesian and parametric forms of the parabola $y^2=4ax$ and rectangular hyperbola $xy=c^2$
- Apply the focus-directrix property of the parabola to solve locus and verification problems
- Calculate equations of tangents and normals to both curves using only Cartesian differentiation
- Solve exam-style questions involving intersections of tangents/normals with curves and coordinate axes

## Standard Forms of Parabolas and Rectangular Hyperbolas

**Standard Conic Forms for FP1** — The only two conic sections examinable in FP1 coordinate systems are the parabola and rectangular hyperbola, with fixed standard and parametric forms given in the formula book.

*Example:* Parabola: Cartesian $y^2=4ax$, parametric $(at^2, 2at)$; Rectangular hyperbola: Cartesian $xy=c^2$, parametric $(ct, \frac{c}{t})$.

**Worked example:** State the coordinates of the point on the parabola $y^2=12x$ corresponding to parameter t=2, and write the Cartesian equation of a rectangular hyperbola with c=4.

1. Step 1: For the parabola, compare $y^2=12x$ to standard $y^2=4ax$ to find $4a=12$, so $a=3$.
2. Step 2: Use parametric coordinates for parabola: $(at^2, 2at) = (3 \times 2^2, 2 \times 3 \times 2) = (12, 12)$.
3. Step 3: For the rectangular hyperbola, substitute $c=4$ into $xy=c^2$: $xy=16$.
4. Final answer: Parabola point = (12,12), hyperbola equation $xy=16$.

Note that the parameter $t$ is only used to define general points on the curves in FP1: you will not be required to manipulate parametric equations beyond substituting values or using them to define points for tangent/normal calculations.

*Calculator:* allowed

## Focus-Directrix Property of the Parabola

**Focus and Directrix of $y^2=4ax$** — The parabola $y^2=4ax$ has a focus at point $(a, 0)$ and a directrix line given by $x=-a$. All points on the parabola are equidistant from this focus and directrix.

*Example:* For $y^2=8x$, $a=2$, so focus is (2,0) and directrix is $x=-2$.

**Worked example:** A point P lies on the parabola $y^2=20x$. Show that the distance from P to the focus equals its perpendicular distance to the directrix for the point P with y-coordinate 10.

1. Step 1: Find $a$ for $y^2=20x$: $4a=20$, so $a=5$. Focus is (5, 0), directrix $x=-5$.
2. Step 2: Find coordinates of P: substitute $y=10$ into $y^2=20x$: $100=20x \rightarrow x=5$, so $P=(5,10)$.
3. Step 3: Calculate distance from P to focus: $\sqrt{(5-5)^2 + (10-0)^2} = \sqrt{100} = 10$.
4. Step 4: Calculate perpendicular distance from P to directrix $x=-5$: horizontal distance since directrix is vertical: $5 - (-5) = 10$.
5. Step 5: Both distances equal 10, so the property holds for point P.

> **tip**
>
> The perpendicular distance from any point $(x,y)$ on the parabola to the directrix $x=-a$ is always $x+a$, since the directrix is vertical. This shortcut saves you from using the full distance formula in exams.

*Calculator:* allowed

## Tangents to Parabolas and Rectangular Hyperbolas

To find the equation of a tangent to either curve, you must differentiate the Cartesian form of the curve to find the gradient of the tangent at the given point, then use the straight line equation $y - y_1 = m(x - x_1)$. Remember parametric differentiation is not permitted here.

**Worked example:** Find the equation of the tangent to the rectangular hyperbola $xy=9$ at the point where $x=3$. Give your answer in the form $ax + by + c = 0$ where $a, b, c$ are integers.

1. Step 1: Rearrange hyperbola equation to Cartesian form for differentiation: $y = \frac{9}{x} = 9x^{-1}$.
2. $$\frac{dy}{dx} = -9x^{-2} = -\frac{9}{x^2}$$
3. Step 2: Find coordinates of the point: when $x=3$, $y=9/3=3$, so point is (3,3).
4. Step 3: Calculate tangent gradient at $x=3$: $m = -\frac{9}{3^2} = -1$.
5. Step 4: Substitute into straight line equation: $y - 3 = -1(x - 3)$
6. Step 5: Rearrange to required form: $y - 3 = -x + 3 \rightarrow x + y - 6 = 0$.

> **exam_tip**
>
> Always rearrange the curve to $y$ as a function of $x$ before differentiating, as this is the method Edexcel expects for FP1, and avoids errors with implicit differentiation which is not required here.

*Calculator:* allowed

## Normals to Parabolas and Rectangular Hyperbolas

The normal to a curve at a point is perpendicular to the tangent at that point, so its gradient is the negative reciprocal of the tangent gradient. Follow the same process as for tangents, but swap the gradient to $m_{normal} = -1/m_{tangent}$ after calculating the tangent gradient.

**Worked example:** Find the equation of the normal to the parabola $y^2=16x$ at the point (1, 4). Give your answer in the form $y = mx + c$.

1. Step 1: Rearrange parabola equation to Cartesian form for differentiation (take positive root as $y=4$ is positive): $y = 4\sqrt{x} = 4x^{1/2}$.
2. $$\frac{dy}{dx} = 4 \times \frac{1}{2}x^{-1/2} = \frac{2}{\sqrt{x}}$$
3. Step 2: Calculate tangent gradient at $x=1$: $m_{tangent} = 2/\sqrt{1} = 2$.
4. Step 3: Calculate normal gradient: $m_{normal} = -1/2$.
5. Step 4: Substitute point (1,4) and normal gradient into straight line equation: $y - 4 = -\frac{1}{2}(x - 1)$
6. Step 5: Rearrange to $y = mx + c$: $y = -\frac{1}{2}x + \frac{1}{2} + 4 = -\frac{1}{2}x + \frac{9}{2}$.

> **warning**
>
> Always take the correct sign of the square root when rearranging $y^2=4ax$ for differentiation: use the sign matching the y-coordinate of the point you are working with, to avoid calculating the wrong gradient.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using parametric differentiation to find tangent/normal gradients
  - Why it fails: Edexcel FP1 explicitly forbids parametric differentiation for this topic, and you will lose method marks even if your final answer is correct
  - Correct: Always differentiate the Cartesian form of the curve ($y$ as a function of $x$) to calculate gradients
- **Wrong:** Using focus and directrix of the rectangular hyperbola to solve problems
  - Why it fails: These are explicitly marked as non-examinable in the FP1 formula book, and questions will never require them for FP1 assessments
  - Correct: Only use the focus-directrix property for the parabola $y^2=4ax$ in FP1 coordinate systems questions
- **Wrong:** Taking the wrong sign of the square root when rearranging $y^2=4ax$ for differentiation
  - Why it fails: This will give you the negative of the correct tangent gradient, leading to incorrect tangent and normal equations
  - Correct: Use the sign of the y-coordinate of the point you are working with when taking the square root of $4ax$
- **Wrong:** Using general hyperbola $x^2/a^2 - y^2/b^2 = 1$ or ellipse equations in answers
  - Why it fails: These are part of FP3 content, not FP1, and will not be accepted as valid answers for FP1 coordinate systems questions
  - Correct: Only use standard forms $y^2=4ax$ for parabolas and $xy=c^2$ for rectangular hyperbolas in FP1
- **Wrong:** Forgetting that the normal gradient is the negative reciprocal of the tangent gradient
  - Why it fails: This will result in a line parallel to the tangent instead of perpendicular, costing you accuracy marks
  - Correct: Always calculate $m_{normal} = -1 / m_{tangent}$ before substituting into the straight line equation for normals

## Cheatsheet

| Concept | Standard Form / Formula | Notes |
| --- | --- | --- |
| Parabola Cartesian | $y^2 = 4ax$ | $a > 0$, opens to the right |
| Parabola Parametric | $(at^2, 2at)$ | $t$ is any real number |
| Parabola Focus | $(a, 0)$ | Given in FP1 formula book |
| Parabola Directrix | $x = -a$ | Given in FP1 formula book |
| Rectangular Hyperbola Cartesian | $xy = c^2$ | $c > 0$, asymptotes at $x=0$, $y=0$ |
| Rectangular Hyperbola Parametric | $(ct, c/t)$ | $t \neq 0$ |
| Tangent Gradient | $m_{tan} = \frac{dy}{dx}$ (Cartesian form) | No parametric differentiation allowed |
| Normal Gradient | $m_{norm} = -1/m_{tan}$ | Perpendicular to tangent |

## What's next

Mastering FP1 coordinate systems is a foundational skill for further conic sections content in FP3, where you will explore general hyperbolas, ellipses, and their eccentricity properties. It also pairs closely with FP1 locus problems, which often ask you to find the path of intersections of tangents and normals to these curves. Before moving on, make sure you can consistently solve full exam-style questions involving tangents, normals, and the focus-directrix property, as these are high-frequency 5-7 mark questions in every FP1 paper. You should also practice combining this content with algebraic manipulation skills from P1 and P2 to simplify equations of lines and solve for intersection points efficiently.

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