Further coordinate systems

Edexcel International A-Level Further Mathematics· FP3 2.1-2.4, WFM03 2018 spec Issue 3· 25 min read

1. Standard Equations of Ellipses and Hyperbolas★★☆☆☆⏱ 5 min

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The two new conic sections assessed in FP3 are the ellipse and general hyperbola, both with horizontal major axes aligned to the x-axis for all exam questions.

📘 Definition

Standard Ellipse

Has Cartesian equation $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ where $a > b$ , and parametric equations $x = a cos t$ , $y = b sin t$ derived using the Pythagorean identity $cos^{2} t + sin^{2} t = 1$ .

📘 Definition

Standard Hyperbola

Has Cartesian equation $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ , with two valid parametric forms: $x = a sec t$ , $y = b tan t$ (uses $t e x t sec^{2} t - tan^{2} t = 1$ , covers both branches) or $x = a cosh t$ , $y = b sinh t$ (uses $t e x t cos h^{2} t - sinh^{2} t = 1$ , covers only the right branch where $x \geq a$ ).

📐 Worked Example

Find the Cartesian equation of the conic with parametric equations $x = 3 sec t$ , $y = 5 tan t$ , and state if it is an ellipse or hyperbola.

1
Rearrange the parametric equations to isolate the trigonometric functions: $\frac{x}{3} = sec t$ , $\frac{y}{5} = tan t$
2
Substitute into the trigonometric identity $sec^{2} t - tan^{2} t = 1$ :
3
$\left(\frac{x}{3}\right)^2 - \left(\frac{y}{5}\right)^2 = 1$
4
Simplify to get the Cartesian equation: $\frac{x ^{2}}{9} - \frac{y ^{2}}{25} = 1$ . This is a hyperbola.

Exam tip:

Use the hyperbolic parametric form for hyperbola problems restricted to the right branch ( $x > 0$ ) to avoid issues with negative secant values.

2. Eccentricity, Foci and Directrices★★★☆☆⏱ 6 min

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All conic sections can be defined by their eccentricity $e$ , which determines their shape. The formula relating $a$ , $b$ and $e$ , plus coordinates of foci and equations of directrices are given in the formula booklet, but you must be able to apply them correctly.

Conic	Eccentricity range	$b^{2}$ relation to $a, e$	Foci coordinates	Directrix equations	Asymptotes
Ellipse	$e < 1$	$b^{2} = a^{2} (1 - e^{2})$	$(\pm a e, 0)$	$x = \pm \frac{a}{e}$	None
Hyperbola	$e > 1$	$b^{2} = a^{2} (e^{2} - 1)$	$(\pm a e, 0)$	$x = \pm \frac{a}{e}$	$y = \pm \frac{b}{a} x$

📐 Worked Example

An ellipse has equation $\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1$ . Calculate its eccentricity, coordinates of the foci, and equations of the directrices.

1
Identify $a$ and $b$ from the standard equation: $a^{2} = 25 ⟹ a = 5$ , $b^{2} = 9 ⟹ b = 3$
2
Substitute into the ellipse eccentricity formula: $9 = 25 (1 - e^{2})$
3
$1 - e^2 = \frac{9}{25} \implies e^2 = \frac{16}{25} \implies e = \frac{4}{5}$
4
Calculate foci coordinates: $\pm a e = \pm 5 \times \frac{4}{5} = \pm 4$ , so foci are $(4, 0)$ and $(- 4, 0)$
5
Calculate directrix equations: $\pm \frac{a}{e} = \pm \frac{5}{4/5} = \pm \frac{25}{4}$ , so directrices are $x = 6.25$ and $x = - 6.25$

Exam tip:

For ellipses, $e < 1$ so $a e < a$ (foci lie inside the ellipse) and $a / e > a$ (directrices lie outside the ellipse). Use this to sanity check your answers.

3. Tangents and Normals to Conics★★★★☆⏱ 8 min

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You can find the gradient of tangents and normals using either implicit differentiation (for Cartesian equations) or parametric differentiation (for parametric coordinates). The tangency condition for a line $y = m x + c$ to touch the conic must be memorized, as it is not given in the formula booklet.

Ellipse tangency condition: $c^{2} = a^{2} m^{2} + b^{2}$
Hyperbola tangency condition: $c^{2} = a^{2} m^{2} - b^{2}$

📐 Worked Example

Find the equation of the tangent to the hyperbola $\frac{x ^{2}}{16} - \frac{y ^{2}}{9} = 1$ at the point where $t = \frac{π}{4}$ (use the parametric form $x = 4 sec t$ , $y = 3 tan t$ ).

1
Find the coordinates of the point at $t = \frac{π}{4}$ : $sec (π /4) = 2$ , $tan (π /4) = 1$ , so $x = 42$ , $y = 3$
2
Differentiate parametrically to find the gradient: $\frac{d x}{d t} = 4 sec t tan t$ , $\frac{d y}{d t} = 3 sec^{2} t$
3
$\frac{dy}{dx} = \frac{3 \text{sec}^2 t}{4 \sec t \tan t} = \frac{3 \sec t}{4 \tan t} = \frac{3}{4 \sin t}$
4
Substitute $t = \frac{π}{4}$ : $\frac{d y}{d x} = \frac{3}{4 \times ( 2 /2 )} = \frac{3}{2 2}$
5
Use point-gradient form for the tangent line: $y - 3 = \frac{3}{2 2} (x - 42)$
6
Simplify to standard form: $3 x - 22 y = 62$

Exam tip:

Always verify your tangent equation using the tangency condition to catch arithmetic errors: for the example above, $c = - 3$ , $c^{2} = 9$ , and $a^{2} m^{2} - b^{2} = 16 \times (9/8) - 9 = 9$ , which matches.

4. Simple Loci Problems★★★★☆⏱ 7 min

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A locus is the set of all points that satisfy a given condition (e.g. midpoint of a tangent, intersection of two normals). The easiest method for most FP3 loci problems is to use parametric coordinates for the point on the conic, express the coordinates of the locus point in terms of the parameter, then eliminate the parameter to get a Cartesian equation.

📐 Worked Example

Find the locus of the midpoint of the line segment joining the focus $(a e, 0)$ of the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ to any point $P$ on the ellipse.

1
Let $P$ have parametric coordinates $(a cos t, b sin t)$ , and let the midpoint be $M (h, k)$
2
Write the midpoint coordinates in terms of $t$ : $h = \frac{a c o s t + a e}{2}$ , $k = \frac{b s i n t + 0}{2}$
3
Rearrange to isolate the trigonometric functions: $\frac{2 h - a e}{a} = cos t$ , $\frac{2 k}{b} = sin t$
4
Use the Pythagorean identity $cos^{2} t + sin^{2} t = 1$ to eliminate $t$ :
5
$\left(\frac{2h - ae}{a}\right)^2 + \left(\frac{2k}{b}\right)^2 = 1$
6
Replace $h$ with $x$ and $k$ with $y$ to get the final locus equation: $\frac{( 2 x - a e ) ^{2}}{a ^{2}} + \frac{4 y ^{2}}{b ^{2}} = 1$ , which is another ellipse.

Exam tip:

Always use parametric coordinates for loci problems where possible, as they reduce the number of variables you need to manipulate compared to using Cartesian coordinates directly.

5. Common Pitfalls

Wrong move:

Using the ellipse eccentricity formula $b^{2} = a^{2} (1 - e^{2})$ for hyperbolas

Why:

Hyperbolas have $e > 1$ , so this would give a negative value for $b^{2}$ , which is impossible

Correct move:

Remember the sign flips for hyperbolas: $b^{2} = a^{2} (e^{2} - 1)$

Wrong move:

Using the ellipse tangency condition $c^{2} = a^{2} m^{2} + b^{2}$ for hyperbolas

Why:

The hyperbola tangency condition has a minus sign, so using the wrong sign will give an invalid value for $c$

Correct move:

Memorize the difference: ellipse uses +, hyperbola uses - for the tangency condition

Wrong move:

Swapping foci and directrix formulae: using $x = \pm a e$ for directrices and $\pm a / e$ for foci

Why:

This gives coordinates/equations that are too large or small, leading to lost method marks

Correct move:

Sanity check with ellipse $e < 1$ : $a e < a$ so foci are inside the ellipse, $a / e > a$ so directrices are outside

Wrong move:

Only using the sec-tan parametric form for hyperbolas even when the problem is restricted to the right branch

Why:

The sec-tan form can produce negative $x$ values that are outside the problem scope, leading to invalid solutions

Correct move:

Use the hyperbolic parametric form $x = a cosh t$ , $y = b sinh t$ for problems restricted to the right branch of the hyperbola

Wrong move:

Forgetting that hyperbolas have asymptotes, while ellipses do not

Why:

Asymptotes are often required for hyperbola problems, and omitting them will lose marks

Correct move:

Recall that hyperbola asymptotes are $y = \pm \frac{b}{a} x$ , given in the formula booklet

6. Quick Reference Cheatsheet

Conic	Standard Cartesian	Parametric Equations	Eccentricity	Foci	Directrices	Tangency Condition $y = m x + c$	Asymptotes
Ellipse ( $a > b$ )	$\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$	$x = a cos t$ , $y = b sin t$	$e < 1$ , $b^{2} = a^{2} (1 - e^{2})$	$(\pm a e, 0)$	$x = \pm \frac{a}{e}$	$c^{2} = a^{2} m^{2} + b^{2}$	None
Hyperbola	$\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$	$x = a sec t, y = b tan t$ / $x = a cosh t, y = b sinh t$	$e > 1$ , $b^{2} = a^{2} (e^{2} - 1)$	$(\pm a e, 0)$	$x = \pm \frac{a}{e}$	$c^{2} = a^{2} m^{2} - b^{2}$	$y = \pm \frac{b}{a} x$

7. Frequently Asked

Do I need to memorize eccentricity formulae for ellipses and hyperbolas?

No, these are provided in the official FP3 formula booklet, but you must be able to apply them correctly to calculate $e$ , foci coordinates and directrix equations.

Is parametric differentiation allowed for finding tangents and normals?

Yes, parametric differentiation is explicitly permitted in FP3, unlike FP1 where it was restricted for conic sections. It is often the fastest method for parameterized points.

Going deeper

official_documentEdexcel FP3 Formula BookletContains all given conic section formulae for the exam

What's Next

Now that you have mastered Further Coordinate Systems for FP3, you can apply these skills to other areas of the Edexcel IAL Further Maths specification. Conic sections often appear alongside differentiation and integration problems, so solidifying your understanding of these coordinate systems will help you tackle more complex calculus questions in FP3. The next topic in FP3 is Further Matrix Algebra, which builds on FP1 matrix content to cover eigenvalues, eigenvectors, and diagonalization, with applications to transformations of conic sections. You should also practice past paper questions on this topic to familiarize yourself with the exam style, as questions often combine multiple parts (eccentricity, tangents, loci) in a single 8-12 mark question. Make sure you can recall the tangency conditions without reference to notes, as these are not provided in the formula book.

→FP1 Coordinate Systems Recap