Further Mathematics · Pearson Edexcel 9FM0 · Option FP1 · Topic 4

Coordinate systems

Conic sections: the parabola and rectangular hyperbola in Year 12, then the ellipse and hyperbola in Year 13. You use their Cartesian and parametric equations, tangents, normals, foci and directrices.

6 sub-topics
4 in Year 12, 2 in Year 13
81 practice questions
6 animated lessons

Practice questions (81) Lessons Worksheets All Edexcel Further Mathematics topics

What the specification says

Every sub-topic Pearson Edexcel can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

4.1Conics: parabola and rectangular hyperbolaY12
4.1Conics: ellipse and hyperbolaY13
4.2Focus–directrix propertiesY12
4.2Eccentricity, foci and directrices of ellipse and hyperbolaY13
4.3Tangents and normals to conicsY12
4.4Loci problemsY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 81 questions in this topic, free, no account needed.

Coordinate systems

Easy

Parabola · Year 12 · 2 marks

y^2 = 12x

State the focus and the directrix.
(2)

Check your answer before you look at the solution

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$y^{2} = 4$ ax has focus $(a, 0)$ . $(3, 0)$
$x = -3$

Total2

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Coordinate systems

Not needed for Year 12Medium

Foci and directrices of a hyperbola · Year 13 · 8 marks

The hyperbola

H

has equation

\dfrac{x^2}{16}

{}- \dfrac{y^2}{9}

{}= 1

.

(a)
Find the eccentricity of $H$ .
(2)
(b)
Find the coordinates of the foci and the equations of the directrices of $H$ .
(3)
(c)
The point $P$ lies on $H$ , above the $x$ -axis, with $x = 5$ . Verify that the distance from $P$ to the focus $S(5, 0)$ is $e$ times the distance from $P$ to the directrix $x = \frac{16}{5}$ .
(3)

Check your answer before you look at the solution

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Part (a)2 marks

From the formulae booklet. $b^2 = a^2(e^2 - 1)$
$e = \frac{5}{4}$

Part (b)3 marks

$(\pm ae, 0) = (\pm 5, 0)$
$x = \pm\frac{a}{e}$
$x = \pm \frac{16}{5}$

Part (c)3 marks

Substitute $x = ae$ into the equation of $H$ . $P\left(5, \frac{9}{4}\right), \ PS$ ${}= \frac{9}{4}$
The distance to the directrix is $|a/e - ae|$ . $e \times \frac{9}{5}$ ${}= \frac{9}{4}$ ${}= PS$

Total8

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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Coordinate systems

Hard

The normal to a parabola · Year 12 · 7 marks

Figure 1 shows the parabola

C

with equation

y^2 = 4x

. The point

P(4, -4)

lies on

C

. The normal to

C

at

P

meets

C

again at the point

Q

.

Figure 1

(a)
Show that the normal to $C$ at $P$ has equation $y - 2x = -12$ .
(4)
(b)
Find the coordinates of $Q$ .
(3)

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Part (a)4 marks

Implicit differentiation, or use x = at², y = 2at with t = -2. $2y\frac{\mathrm{d}y}{\mathrm{d}x}$ ${}= 4$ ${}\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}$ ${}= \frac{2}{y}$ ${}= - \frac{1}{2}$
The normal is perpendicular: $m_{1}m_{2} = -1$ . $m_{\text{normal}} = 2$
$y - y_{1} = m(x - x_{1})$ . $y - 2x = -12$

Part (b)3 marks

Substitute $x = y^{2}/4$ into the normal. $\left(\frac{y^2}{4}\right) \text{ in the normal}: \ -\frac{1}{2}y^2$ ${}+ y$ ${}+ 12$ ${}= 0$
One root is already known. $y = -4 \text{ (at P) or } y = 6$
$Q(9, 6)$

Total7

MMethod mark: for a correct method, even with a slip.
AAccuracy mark: for a correct answer from a correct method.
BIndependent mark: for a correct fact or result on its own, with no method needed.

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