Edexcel IAL Maths FP3.2 Coordinate Systems

[Maximum number: 10]

The ellipse E has equation

\frac{x^2}{9}+\frac{y^2}{4}=1

The line l has equation y=kx-3, where k is a constant.
Given that E and l meet at 2 distinct points P and Q,

(a)

show that the x coordinates of P and Q are solutions of the equation

(9k^2+4)x^2-54kx+45=0

The point M is the midpoint of PQ.

[ 2 ]

\frac{x^2}{9}+\frac{(kx-3)^2}{4}=1

4x^2+9(k^2x^2-6kx+9)=36

(9k^2+4)x^2-54kx+45=0

M1 A1*

(b)

Determine, in simplest form in terms of k, the coordinates of M.

[ 3 ]

Using the sum of roots,

x_M=\frac12\left(\frac{54k}{9k^2+4}\right)=\frac{27k}{9k^2+4}

Then

y_M=k\left(\frac{27k}{9k^2+4}\right)-3=\frac{27k^2-27k^2-12}{9k^2+4}=-\frac{12}{9k^2+4}

M\left(\frac{27k}{9k^2+4},-\frac{12}{9k^2+4}\right)

M1 dM1 A1

(c)

Hence show that, as k varies, M lies on the curve with equation

x^2+py^2=qy

where p and q are constants to be determined.

[ 5 ]

From part (b),

x^2=\frac{729k^2}{(9k^2+4)^2},\qquad y^2=\frac{144}{(9k^2+4)^2}

x^2+py^2=\frac{729k^2+144p}{(9k^2+4)^2}

Since

y=-\frac{12}{9k^2+4}

match

\frac{729k^2+144p}{(9k^2+4)^2}=-\frac{12q}{9k^2+4}

This gives

p=\frac94,\qquad q=-\frac{27}{4}

Hence

x^2+\frac94y^2=-\frac{27}{4}y

M1 dM1 A1 ddM1 A1

[Maximum number: 13]

The ellipse E has equation

\frac{x^{2}}{9}+\frac{y^{2}}{4}=1

(a)

Determine the eccentricity of E

[ 2 ]

$b^{2}=a^{2}\left(1-e^{2}\right) \Rightarrow 4=9\left(1-e^{2}\right) \Rightarrow e=\ldots$

or e.g.
$e=\sqrt{1-\frac{b^{2}}{a^{2}}} \Rightarrow e=\ldots$

Uses a correct formula with a and b
correctly placed to find a value for e

M1

$e=\frac{\sqrt{5}}{3}$

Correct value (or equivalent)
$e= \pm \frac{\sqrt{5}}{3} \text { scores A0 }$

A1

(2)

(b)

Hence, for this ellipse, determine

[ 2 ]

(i)

the coordinates of the foci,

[ 1 ]

$( \pm a e, 0)=( \pm \sqrt{5}, 0) \text { or }\left( \pm 3 \frac{\sqrt{5}}{3}, 0\right)$

Correct foci. Must be coordinates but allow unsimplified and isw if necessary.

Follow through their e so allow for ( ± 3× their e, 0 )

B1ft

(ii)

the equations of the directrices.

The point P lies on E and has coordinates $(3 \cos \theta, 2 \sin \theta)$ .

The line $l_{1}$ is the tangent to E at the point P

[ 1 ]

$x= \pm \frac{a}{e}= \pm \frac{9}{\sqrt{5}} \text { or } x= \pm \frac{3}{\frac{\sqrt{5}}{3}}$

Correct directrices. Must be equations but allow unsimplified and isw if necessary.

Follow through their e so allow for $x= \pm 3$ /their e

B1ft

(2)

Special case:

Use of $a^{2}$ for a and $b^{2}$ for b consistently scores M0A0 in (a) and B1ft B1ft in (b)

This gives $e=\frac{\sqrt{65}}{9},( \pm \sqrt{65}, 0), x= \pm \frac{81}{\sqrt{65}}$

(c)

Using calculus, show that an equation for $l_{1}$ is

2 x \cos \theta+3 y \sin \theta=6

The line $l_{2}$ passes through the origin and is perpendicular to $l_{1}$
The line $l_{1}$ intersects the line $l_{2}$ at the point Q

[ 3 ]

For $(x,y)=(3\cos\theta,2\sin\theta)$ ,
$\frac{dx}{d\theta}=-3\sin\theta,\qquad \frac{dy}{d\theta}=2\cos\theta.$
Thus
$\frac{dy}{dx}=\frac{2\cos\theta}{-3\sin\theta}.$
The tangent is
$y-2\sin\theta=\frac{2\cos\theta}{-3\sin\theta}(x-3\cos\theta),$
which rearranges to
$2x\cos\theta+3y\sin\theta=6.$

M1 M1 A1

(d)

Determine the coordinates of Q

[ 3 ]

$l_{2}: y=\frac{3 \sin \theta}{2 \cos \theta} x$

Correct equation for $l_{2}$

B1 ${ }^{\text {Bllas }} \mathrm{Cl}_{\mathrm{Cl}}$ $2 x \cos \theta+3 y \sin \theta=6, y=\frac{3 \sin \theta}{2 \cos \theta} x$ $\Rightarrow x=\ldots, y=\ldots$

Complete method for Q

M1

$Q:\left(\frac{12 \cos \theta}{4 \cos ^{2} \theta+9 \sin ^{2} \theta}, \frac{18 \sin \theta}{4 \cos ^{2} \theta+9 \sin ^{2} \theta}\right)$

Correct coordinates. Allow as $x=\ldots, y=\ldots$ and allow equivalent correct expressions as long as they are single fractions
$\text { e.g. } x=\frac{12 \cos \theta}{4+5 \sin ^{2} \theta} \quad y=\frac{18 \sin \theta}{4+5 \sin ^{2} \theta}, \quad x=\frac{12 \cos \theta}{9-5 \cos ^{2} \theta} \quad y=\frac{18 \sin \theta}{9-5 \cos ^{2} \theta}$

A1

(3)

(e)

Show that, as $\theta$ varies, the point Q lies on the curve with equation

\left(x^{2}+y^{2}\right)^{2}=\alpha x^{2}+\beta y^{2}

where $\alpha$ and $\beta$ are constants to be determined.

[ 3 ]

For the point Q,
$\frac{y}{x}=\frac32\tan\theta.$
Using the result from the preceding parts to eliminate $\theta$ gives
$(x^2+y^2)^2=9x^2+4y^2.$
Therefore
$\alpha=9,\qquad \beta=4.$

M1 M1 A1

Pearson Edexcel IAL Mathematics FP3.2 Further coordinate systems Question Bank

Exam points

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Covered syllabus points

FP3.2 - Further coordinate systems question 1

Question (a)

Question (b)

Question (c)

FP3.2 - Further coordinate systems question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question (d)

Question (e)