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WFM03 · Chapter 2
Further Coordinate Systems
Ellipses, hyperbolas and parabolas are all the same family of curve
viewed from different angles — and once you can write them parametrically,
you can find tangents, normals and loci using the same chain-rule trick
every single time.
Summary
Ellipse: Cartesian and parametric forms, foci, directrices, eccentricity < 1, and the "string method" definition PS + PS′ = 2a.
Hyperbola: the "twin" of the ellipse with eccentricity > 1, asymptotes, and the same focus-directrix definition PS = ePN.
Parabola: the special case where e = 1 — one focus, one directrix, no centre of symmetry.
Parametric differentiation: finding dy/dx from x(θ) and y(θ) using the chain rule, without ever eliminating the parameter.
Tangents and normals: using the parametric gradient to write down the equation of a tangent or normal at a general point.
Finding a locus: tracking a point that depends on a parameter, then eliminating the parameter to get a Cartesian equation for the path it traces.
The Ellipse
Think of an ellipse as a circle that's been squashed. Instead of one radius, it has two
"half-widths" — a along the x-axis and
b along the y-axis.
Cartesian & Parametric Forms
Cartesian: x²/a² + y²/b² = 1 | Parametric: x = a cos θ, y = b sin θ
Every point on the ellipse can be written as (a cos θ, b sin θ) for some angle θ — this is just because cos²θ + sin²θ = 1 automatically satisfies the Cartesian equation.
Foci, directrices and eccentricity
An ellipse has two foci (special interior points) at S(ae, 0) and S′(−ae, 0),
and two directrices (fixed vertical lines) at x = ±a/e. The
eccentricity e tells you how "squashed" the ellipse is — for an ellipse,
always e < 1. The three quantities a, b and e are linked by:
Key relationship
b² = a²(1 − e²)
The focus–directrix definition
Here's the deep idea underneath all of this: an ellipse is the set of all points P such that
the distance to a focus is a fixed fraction of the distance to the corresponding directrix:
Definition
PS = e · PN
where S is a focus, N is the foot of the perpendicular from P to the corresponding directrix, and this holds true for either focus with its matching directrix.
🧵 Why the "string method" works
If you pin a loop of string around both foci and trace it taut with a pencil, you draw a
perfect ellipse. Here's the proof buried in the notes: since PS = ePN and PS′ = ePN′, adding
these gives PS + PS′ = e(PN + PN′) = e·NN′ = e·(2a/e) = 2a. So the sum of distances from
any point on the ellipse to the two foci is always 2a — a constant. That
constant string length is exactly 2a.
Worked Example
Q: The ellipse x²/25 + y²/16 = 1 has eccentricity e. Find e, and the coordinates of the foci.
Step 3: Foci are at (±ae, 0) = (±5 × 3/5, 0) = (±3, 0).
Practice Question
An ellipse has parametric equations x = 6 cos θ, y = 2 sin θ. Find its eccentricity and the equations of its directrices.
The Hyperbola
The hyperbola is the ellipse's more dramatic cousin — instead of a closed loop, it's
two open branches that shoot off towards infinity, hugging straight-line
"guide rails" called asymptotes as they go.
Cartesian & Parametric Forms
Cartesian: x²/a² − y²/b² = 1 | Parametric: x = a cosh θ, y = b sinh θ
This works because cosh²θ − sinh²θ = 1 — the hyperbolic identity that mirrors cos²θ + sin²θ = 1 for the ellipse. (You could also use x = a sec θ, y = b tan θ, since sec²θ − tan²θ = 1 too — both are valid parametrisations.)
Asymptotes
As x gets very large, the curve gets closer and closer to two straight lines through the
origin, but never quite touches them:
Asymptotes
x/a = ±y/b (i.e. y = ±(b/a)x)
Foci, directrices, eccentricity
Just like the ellipse, the hyperbola has foci S(ae, 0) and S′(−ae, 0), and directrices
at x = ±a/e — but this time the eccentricity is always greater than 1,
and the relationship between a, b and e flips a sign:
Key relationship
b² = a²(e² − 1)
The focus–directrix definition is identical in form to the ellipse's — PS = e·PN —
the only thing that changes is that e > 1 instead of e < 1. That single number is what
decides whether you get a closed ellipse or an open hyperbola.
💡 Foci on the y-axis instead?
If the equation is written as y²/c² − x²/d² = 1, it's exactly the same shape but rotated —
now the foci sit on the y-axis instead of the x-axis. Don't panic if you see this form;
just swap the roles of x and y in everything you know.
Worked Example
Q: Find the eccentricity and asymptotes of the hyperbola x²/9 − y²/16 = 1.
A hyperbola has a = 2 and eccentricity e = 2. Find b² and state the coordinates of the foci.
The Parabola
The parabola is the "boundary case" between the ellipse and the hyperbola — it happens
exactly when e = 1. It only has one focus and one directrix (no second one,
because there's no second branch or second focal point to balance).
Cartesian & Parametric Forms
Cartesian: y² = 4ax | Parametric: x = at², y = 2at
Substituting the parametric form back in confirms it: y² = (2at)² = 4a²t² and 4ax = 4a(at²) = 4a²t². They match for every value of t.
Focus and directrix
Focus S = (a, 0) | Directrix: x = −a
Because e = 1 exactly, the focus-directrix definition simplifies beautifully — there's no
"e" multiplying anything anymore:
Definition
PS = PN
Every point on the parabola is exactly as far from the focus as it is from the directrix.
🔗 The big-picture connection
All three curves are governed by one single idea: PS = e·PN. As e sweeps from 0 up through 1
and beyond, the shape morphs smoothly: e = 0 gives a circle, 0 < e < 1 gives an ellipse,
e = 1 gives a parabola, and e > 1 gives a hyperbola. It's genuinely one family, not three
unrelated curves to memorise separately.
Worked Example
Q: A parabola has equation y² = 12x. State the focus and directrix, and find the coordinates of the point where t = 3 on the parametric form.
Step 3: Parametric point at t = 3: x = at² = 3(9) = 27, y = 2at = 2(3)(3) = 18. Point is (27, 18).
Practice Question
Find the Cartesian equation of the parabola whose parametric equations are x = 5t², y = 10t.
Parametric Differentiation
Once a curve is written parametrically — x and y both given in terms of some parameter
θ or t — you almost never want to eliminate the parameter just to differentiate. Instead,
the chain rule hands you a shortcut for free.
The Rule
dy/dx = (dy/dθ) ÷ (dx/dθ)
This comes straight from the chain rule: dy/dθ = (dy/dx)(dx/dθ), so rearranging gives dy/dx = (dy/dθ)/(dx/dθ). Differentiate y and x separately with respect to the parameter, then divide.
This works for any parameter — θ for the ellipse/hyperbola, t for the parabola —
and it's usually far less painful than trying to make y a function of x directly.
Worked Example
Q: Find dy/dx for the curve x = 6t, y = 6/t, at t = 3 — using two different methods.
Method 1 (parametric): dx/dt = 6, dy/dt = −6t⁻². So dy/dx = (−6t⁻²)/6 = −1/t². At t = 3: dy/dx = −1/9.
Method 2 (eliminate first): xy = 36 ⇒ y = 36/x ⇒ dy/dx = −36/x². At t = 3, x = 18, so dy/dx = −36/324 = −1/9. Same answer — the parametric method just skipped the elimination step.
🎯 When each method wins
Parametric differentiation is faster when eliminating the parameter would be messy
(like with cos θ / sin θ pairs). Implicit differentiation is sometimes faster when the
Cartesian equation is already simple, like xy = constant. Learn to do both, then pick
whichever is quicker for the question in front of you.
Practice Question
Find dy/dx in terms of θ for the ellipse x = 4 cos θ, y = 3 sin θ.
Tangents and Normals
Once you have dy/dx at a general parametric point, finding the tangent or normal is just
the same "point + gradient" line equation you've used since GCSE — the only new skill is
keeping the algebra tidy with a parameter still sitting in there.
Recipe
y − y₁ = m(x − x₁)
where (x₁, y₁) is the parametric point and m = dy/dx evaluated there for the tangent, or m = −1/(dy/dx) for the normal (perpendicular gradient).
Worked Example
Q: Find the equation of the normal to the curve x = 5 cos θ, y = 8 sin θ at θ = π/3.
Step 1: At θ = π/3: cos θ = 1/2, sin θ = √3/2 ⇒ x = 5/2, y = 4√3.
Step 2: dy/dx = (dy/dθ)/(dx/dθ) = (8 cos θ)/(−5 sin θ) = (8 × 1/2)/(−5 × √3/2) = −8/(5√3).
Step 3: Gradient of normal = negative reciprocal = 5√3/8.
Step 4: y − 4√3 = (5√3/8)(x − 5/2) ⇒ multiplying through and tidying: 5√3 x − 8y + 39√3/2 = 0.
⚠️ Easy slip-up
Always double check whether the question wants the tangent or the normal
— they use exactly the same working except for one step (whether you use m or −1/m). Mixing
them up is the single most common way to lose easy marks on this topic.
Practice Question
Find the equation of the tangent to the parabola x = 2t², y = 4t at the point where t = 1.
Finding a Locus
A locus question describes some point (often related to a tangent or a chord) whose
position depends on a parameter as it moves along a curve — and asks you to describe the
path that point traces out. The strategy is always the same three steps:
Write the coordinates of the moving point in terms of the parameter (θ or t).
Call these coordinates X and Y (capital letters, to keep them distinct from the original curve's x, y).
Eliminate the parameter — usually using an identity like cos²θ + sin²θ = 1 — to get a Cartesian equation purely in X and Y.
Worked Example
Q: The tangent to the ellipse x²/9 + y²/16 = 1 at the point P(3 cos θ, 4 sin θ) crosses the x-axis at A and the y-axis at B. Find the locus of the midpoint of AB as θ varies.
Step 1: dy/dx = (dy/dθ)/(dx/dθ) = (4 cos θ)/(−3 sin θ).
Step 2: Tangent equation: y − 4 sin θ = [4cos θ/(−3 sin θ)](x − 3 cos θ). Multiplying by −3 sin θ and simplifying using cos²θ + sin²θ = 1 gives: 3y sin θ + 4x cos θ = 12.
Step 3: At A (y = 0): x = 3/cos θ. At B (x = 0): y = 4/sin θ.
Step 4: Midpoint of AB: X = 3/(2cos θ), Y = 2/sin θ ⇒ cos θ = 3/(2X), sin θ = 2/Y.
Step 5: Use cos²θ + sin²θ = 1: 9/(4X²) + 4/Y² = 1 — this is the equation of the locus.
🧭 The one identity that does all the work
Almost every locus elimination in this chapter comes down to substituting your expressions
for cos θ and sin θ (or cosh θ and sinh θ) into cos²θ + sin²θ = 1 (or cosh²θ − sinh²θ = 1).
Before you touch the algebra, ask: "what am I going to substitute into, at the very end?"
That tells you exactly what form to rearrange X and Y into.
Practice Question
A point Q has coordinates X = 2 + 3 cos θ, Y = −1 + 3 sin θ as θ varies. Find the Cartesian equation of the locus of Q, and describe the shape it represents.
What to Memorise
Curve
Cartesian
Parametric
Key facts
Ellipse
x²/a² + y²/b² = 1
x = a cos θ, y = b sin θ
e < 1, b² = a²(1−e²), foci (±ae,0), directrices x = ±a/e
Hyperbola
x²/a² − y²/b² = 1
x = a cosh θ, y = b sinh θ
e > 1, b² = a²(e²−1), asymptotes y = ±(b/a)x
Parabola
y² = 4ax
x = at², y = 2at
e = 1, focus (a,0), directrix x = −a
Rule
Statement
Focus–directrix definition
PS = e·PN for any of the three curves
Parametric gradient
dy/dx = (dy/dθ) ÷ (dx/dθ)
Tangent gradient
use m = dy/dx at the given point
Normal gradient
use m = −1/(dy/dx) at the given point
Ellipse string property
PS + PS′ = 2a (constant)
📌 Quick sanity check
If you forget which relationship goes with which curve, remember: ellipse is "closed" so
it's a minus in b² = a²(1 − e²); hyperbola is "open" so it's a plus in
b² = a²(e² − 1). Closed → subtract, open → add.
Concepts Checklist
Exam Tips
⚠️Don't mix up e < 1 and e > 1. Examiners love asking you to identify which curve you're dealing with just from the b² relationship — check the sign before assuming it's an ellipse.
⚠️Always differentiate with respect to the parameter, not x. A very common error is trying to differentiate y = b sin θ "with respect to x" directly — you need dy/dθ and dx/dθ separately first.
⚠️Normal vs tangent gradients get swapped under exam pressure. Write "tangent: m" or "normal: −1/m" explicitly before you substitute numbers in — it costs seconds and saves marks.
⚠️In locus questions, use capital X and Y for the coordinates of the new (moving) point. Re-using lowercase x and y that already belong to the original curve is the most common source of algebra errors in this topic.
✅Before eliminating a parameter, stop and ask which identity you're aiming for — cos²θ + sin²θ = 1 for ellipse-type parametrisations, cosh²θ − sinh²θ = 1 for hyperbola-type ones. Rearrange your X and Y expressions to isolate cos θ (or cosh θ) and sin θ (or sinh θ) on their own first.
✅Sketch it if you're unsure. A rough sketch of the ellipse/hyperbola/parabola with the foci and directrices marked on often reveals arithmetic mistakes (e.g. a directrix that ends up inside the curve, which is impossible) before you submit a wrong final answer.
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