Edexcel A-Level Mathematics A2 Fp3 2 Further Coordinate Systems Questions

Practise ellipse and hyperbola work involving parametric forms, foci, directrices, tangents, normals and loci from varying points.

Exam points

use Cartesian or parametric equations to find intersections, midpoints and asymptotes
apply focus-directrix data to calculate foci, directrices, eccentricity or curve equations
eliminate parameters from tangent, midpoint or intersection results to form a locus equation

Covered syllabus points

FP3.2.1 - Cartesian and parametric equations FP3.2.2 - Focus-directrix properties FP3.2.3 - Tangents and normals to these FP3.2.4 - Simple loci problems

[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)

(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

[Maximum number: 13]

The ellipse E has equation

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

(a)

Determine the eccentricity of E

[ 2 ]

$b^2=a^2\left(1-e^2\right)\Rightarrow4=9\left(1-e^2\right)$

$e=\frac{\sqrt5}{3}$

(b)

Hence, for this ellipse, determine

[ 2 ]

(i)

the coordinates of the foci,

[ 1 ]

$(\pm ae,0)=(\pm\sqrt5,0)$

(ii)

the equations of the directrices.

[ 1 ]

$x=\pm\frac{a}{e}=\pm\frac{9}{\sqrt5}$

(c)

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

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

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

[ 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
$2x\cos\theta+3y\sin\theta=6.$

(d)

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

Determine the coordinates of Q

[ 3 ]

$l_2:\ y=\frac{3\sin\theta}{2\cos\theta}x$

$2x\cos\theta+3y\sin\theta=6,\qquad y=\frac{3\sin\theta}{2\cos\theta}x$

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

(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
$\alpha=9,\qquad \beta=4.$

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question (i)

Question (ii)

Question (c)

Question (d)

Question (e)