What are tangent gradient slips?

For the parabola the gradient at parameter t is (1)/(t); recompute it from (dy)/(dx) = (2a)/(y) rather than guessing.

What is wrong eccentricity relation?

Use b^2 = a^2(1 - e^2) for the ellipse and b^2 = a^2(e^2 - 1) for the hyperbola; the sign inside the bracket differs.

Find the eccentricity of the ellipse (x^2)/(16) + (y^2)/(7) = 1. [3 marks]

Pearson Edexcel Edexcel 2021-style practice: The ellipse has equation (x^2)/(25) + (y^2)/(9) = 1. Find its eccentricity, the coordinates of its foci, and the equations of its directrices.

Identify a and b, then use b^2 = a^2(1 - e^2). Here a^2 = 25 so a = 5, and b^2 = 9 (M1). Using b^2 = a^2(1 - e^2): 9 = 25(1 - e^2), so 1 - e^2 = (9)/(25) and e^2 = (16)/(25) (M1 A1), giving e = (4)/(5) (A1). Foci at ( ae, 0) = ( 5 · (4)/(5), 0) = ( 4, 0) (A1). Directrices at x = (a)/(e) = (5)/(4/5) = (25)/(4) (M1 A1).

Pearson Edexcel Edexcel 2023-style practice: The point P(ct, (c)/(t)) lies on the rectangular hyperbola xy = c^2. Find the equation of the tangent to the curve at P.

Differentiate to find the gradient at P, then use point-gradient form. Write y = (c^2)/(x), so (dy)/(dx) = -(c^2)/(x^2) (M1). At x = ct: (dy)/(dx) = -(c^2)/(c^2 t^2) = -(1)/(t^2) (A1). Tangent at (ct, (c)/(t)): y - (c)/(t) = -(1)/(t^2)(x - ct) (M1). Multiply by t^2 and simplify: t^2 y - ct = -x + ct, so x + t^2 y = 2ct (A1 A1).

EnglandFurther MathsSyllabus dot point

How are the conic sections defined, and how do you work with their tangents, normals and parametric forms?

The parabola, ellipse and hyperbola in Cartesian and parametric form, foci and directrices, tangents and normals, and the rectangular hyperbola.

A focused answer to the Edexcel A-Level Further Mathematics Further Pure coordinate systems content, covering the parabola, ellipse and hyperbola in Cartesian and parametric form, foci and directrices, tangents and normals to conics, and the rectangular hyperbola.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The standard conics
Tangents and normals
Focus, directrix and eccentricity
Examples in context
Try this

What this dot point is asking

Edexcel Further Pure wants you to recognise the parabola, ellipse and hyperbola in Cartesian and parametric form, locate foci and directrices, derive equations of tangents and normals (usually most cleanly from the parametric form), and work with the rectangular hyperbola $xy = c^2$ . Conic questions are typically multi-part "show that" derivations, so every algebraic step must be shown.

The standard conics

Each conic has a Cartesian equation and a convenient parametric form. The parametric form is the key to tidy tangent and normal work, because the gradient comes out as a simple expression in the parameter.

Tangents and normals

To find the tangent at a parametric point, differentiate to get $\frac{dy}{dx}$ in terms of the parameter, then use the point-gradient form $y - y_1 = m(x - x_1)$ . The normal uses the negative reciprocal of the tangent gradient.

Tangent to a parabola

Find the tangent to $y^2 = 4ax$ at the point $(at^2, 2at)$ .

Step 1: Differentiate implicitly to find the gradient

Differentiating $y^2 = 4ax$ with respect to $x$ gives $2y\dfrac{dy}{dx} = 4a$ , so:

\frac{dy}{dx} = \frac{2a}{y}.

At the parametric point $(at^2, 2at)$ the $y$ -coordinate is $2at$ , so the gradient there is:

\frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}.

Step 2: Write the tangent using point-gradient form

The tangent passes through $(at^2, 2at)$ with gradient $\dfrac{1}{t}$ :

y - 2at = \frac{1}{t}(x - at^2).

Step 3: Rearrange to the standard form

Multiply through by $t$ :

ty - 2at^2 = x - at^2,

which rearranges to:

ty = x + at^2.

Final answer: the tangent at parameter $t$ is $ty = x + at^2$ .

Normal to the rectangular hyperbola

Find the normal to $xy = c^2$ at $\left(ct, \dfrac{c}{t}\right)$ .

Step 1: Find the tangent gradient by differentiating

Write the curve as $y = \dfrac{c^2}{x}$ , so $\dfrac{dy}{dx} = -\dfrac{c^2}{x^2}$ . At $x = ct$ this becomes:

\frac{dy}{dx} = -\frac{c^2}{c^2 t^2} = -\frac{1}{t^2}.

Step 2: Find the normal gradient

The normal is perpendicular to the tangent, so its gradient is the negative reciprocal of $-\dfrac{1}{t^2}$ , which is $t^2$ .

Step 3: Write the equation of the normal

Using point-gradient form at $\left(ct, \dfrac{c}{t}\right)$ :

y - \frac{c}{t} = t^2(x - ct).

Step 4: Rearrange to the standard form

Multiply through by $t$ to clear the fraction:

ty - c = t^3 x - ct^4,

which rearranges to:

t^3 x - ty = c(t^4 - 1).

Final answer: the normal at parameter $t$ is $t^3 x - ty = c(t^4 - 1)$ .

Focus, directrix and eccentricity

Every conic is the locus of points whose distance to a fixed focus is a constant multiple (the eccentricity $e$ ) of the distance to a fixed directrix line. The value of $e$ classifies the conic, and for the ellipse and hyperbola it determines the positions of the foci and directrices.

Examples in context

Coordinate systems link to several Further Maths threads. The parametric tangent and normal work uses the same point-gradient methods as core differentiation, and the parametric forms reappear when computing arc lengths and areas. Polar coordinates give an alternative description of conics with the focus at the pole, the natural setting for planetary orbits (ellipses with the Sun at a focus, by Kepler's first law). The reflective property of the parabola (rays parallel to the axis reflect through the focus) underlies satellite dishes and headlamps. Vectors and the dot product supply angles between tangents and chords.

Try this

Q1. Write down a parametric point on the rectangular hyperbola $xy = 16$ . [1 mark]

Cue. $c^2 = 16$ so $c = 4$ , giving $\left(4t, \frac{4}{t}\right)$ .

Q2. State the focus and directrix of $y^2 = 12x$ . [2 marks]

Cue. $4a = 12$ so $a = 3$ : focus $(3, 0)$ , directrix $x = -3$ .

Q3. Find the eccentricity of the ellipse $\frac{x^2}{16} + \frac{y^2}{7} = 1$ . [3 marks]

Cue. $7 = 16(1 - e^2)$ gives $e^2 = \frac{9}{16}$ , so $e = \frac{3}{4}$ .

Exam-style practice questions

Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Edexcel 20196 marksThe point

P(at^2, 2at)

lies on the parabola

y^2 = 4ax

. Show that the equation of the normal to the parabola at

P

y + tx = 2at + at^3

Show worked answer →

Find the tangent gradient by differentiating, take the negative reciprocal for the normal, then use point-gradient form.

Differentiating $y^2 = 4ax$ implicitly: $2y\frac{dy}{dx} = 4a$ , so $\frac{dy}{dx} = \frac{2a}{y} = \frac{2a}{2at} = \frac{1}{t}$ at $P$ (M1 A1).

The normal gradient is the negative reciprocal, $-t$ (M1).

Point-gradient form at $(at^2, 2at)$ : $y - 2at = -t(x - at^2)$ (M1). Expanding, $y - 2at = -tx + at^3$ (A1), so $y + tx = 2at + at^3$ as required (A1).

Edexcel 20217 marksThe ellipse has equation

\frac{x^2}{25} + \frac{y^2}{9} = 1

. Find its eccentricity, the coordinates of its foci, and the equations of its directrices.

Show worked answer →

Identify $a$ and $b$ , then use $b^2 = a^2(1 - e^2)$ .

Here $a^2 = 25$ so $a = 5$ , and $b^2 = 9$ (M1). Using $b^2 = a^2(1 - e^2)$ : $9 = 25(1 - e^2)$ , so $1 - e^2 = \frac{9}{25}$ and $e^2 = \frac{16}{25}$ (M1 A1), giving $e = \frac{4}{5}$ (A1).

Foci at $(\pm ae, 0) = (\pm 5 \cdot \frac{4}{5}, 0) = (\pm 4, 0)$ (A1). Directrices at $x = \pm\frac{a}{e} = \pm\frac{5}{4/5} = \pm\frac{25}{4}$ (M1 A1).

Edexcel 20235 marksThe point

P\left(ct, \frac{c}{t}\right)

lies on the rectangular hyperbola

xy = c^2

. Find the equation of the tangent to the curve at

P

Show worked answer →

Differentiate to find the gradient at $P$ , then use point-gradient form.

Write $y = \frac{c^2}{x}$ , so $\frac{dy}{dx} = -\frac{c^2}{x^2}$ (M1). At $x = ct$ : $\frac{dy}{dx} = -\frac{c^2}{c^2 t^2} = -\frac{1}{t^2}$ (A1).

Tangent at $\left(ct, \frac{c}{t}\right)$ : $y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)$ (M1).

Multiply by $t^2$ and simplify: $t^2 y - ct = -x + ct$ , so $x + t^2 y = 2ct$ (A1 A1).

Related dot points

Sources & how we know this

Pearson Edexcel A-Level Further Mathematics (9FM0) specification — Pearson Edexcel (2017)