Focus-directrix properties and eccentricityEdexcel International A Level Further Maths: Revision notes

Section 1

The focus-directrix property and eccentricity

A focus $S$ , a line called the directrix and a number $e>0$ called the eccentricity define a curve. It is the locus of points $P$ for which $PS=e\times PM,$ where $M$ is the foot of the perpendicular from $P$ to the directrix. If $0<e<1$ the curve is an ellipse; if $e>1$ it is a hyperbola. The ellipse and hyperbola each have two foci and two directrices, symmetrical about the centre, and the property holds for each focus with its corresponding directrix (the nearer one on the same side).

Key termsfocusdirectrixeccentricitycorresponding directrix

Common mistake

Using the wrong directrix with a focus. The focus $(ae,0)$ goes with $x=\frac ae$ , and $(-ae,0)$ with $x=-\frac ae$ .

Section 2

The ellipse

For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$ and $0<e<1$ : $b^2=a^2\left(1-e^2\right),\quad\text{foci }(\pm ae,0),\quad\text{directrices }x=\pm\frac ae.$ Since $e<1$ , the foci lie inside the ellipse ( $ae<a$ ) and the directrices outside it ( $\frac ae>a$ ). Example: $\frac{x^2}{25}+\frac{y^2}{16}=1$ gives $16=25\left(1-e^2\right)$ , so $e=\frac35$ , foci $(\pm3,0)$ and directrices $x=\pm\frac{25}{3}$ .

Key termssemi-major axis

Exam tip

Find $e$ first from $b^2=a^2(1-e^2)$ ; the foci and directrices then follow from $ae$ and $\frac ae$ .

Section 3

The hyperbola

For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $e>1$ : $b^2=a^2\left(e^2-1\right),\quad\text{foci }(\pm ae,0),\quad\text{directrices }x=\pm\frac ae.$ Here $ae>a$ so the foci lie inside the branches and $\frac ae<a$ so the directrices lie between the two vertices. Example: $\frac{x^2}{9}-\frac{y^2}{16}=1$ gives $16=9\left(e^2-1\right)$ , so $e=\frac53$ , foci $(\pm5,0)$ and directrices $x=\pm\frac95$ .

Common mistake

Using $1-e^2$ for the hyperbola. It is $e^2-1$ , and $e>1$ .

Section 4

Verifying the property

To verify $PS=e\times PM$ for a point $P$ : find $S$ and the corresponding directrix, work out both distances, and check the ratio. For $P\left(5,\frac{16}{3}\right)$ on $\frac{x^2}{9}-\frac{y^2}{16}=1$ : $S=(5,0)$ gives $PS=\frac{16}{3}$ , the directrix $x=\frac95$ gives $PM=5-\frac95=\frac{16}{5}$ , and $\frac{PS}{PM}=\frac53=e$ . Distance to a vertical line $x=k$ is $|x-k|$ . For the distance from $P$ to the focus use the distance formula.

Exam tip

Check the point lies on the curve first, then use the focus and directrix on the same side.

Section 5

Finding the equation from the property

A locus given by $PS=e\times PM$ can be turned into a Cartesian equation by squaring. For $S(6,0)$ , directrix $x=\frac83$ and $e=\frac32$ : $(x-6)^2+y^2=\frac94\left(x-\frac83\right)^2$ , which gives $x^2-12x+36+y^2=\frac94x^2-12x+16$ , so $y^2=\frac54x^2-20$ and $\frac{x^2}{16}-\frac{y^2}{20}=1$ . Conversely, given $e$ and a focus, use $ae=$ (focus distance) to find $a$ , then $b^2$ from the $e$ relation. For foci $(\pm4,0)$ and $e=\frac23$ : $a=6$ , $b^2=36\left(1-\frac49\right)=20$ .

Key termslocus

Exam tip

Square both sides of $PS=e\times PM$ to remove the square root from $PS$ .

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Focus-directrix properties and eccentricity

The ellipse $E$ has equation $\frac{x^2}{25}+\frac{y^2}{16}=1$ .
Find the equations of the directrices of $E$ .2 marks
The hyperbola $H$ has equation $\frac{x^2}{9}-\frac{y^2}{16}=1$ .
The point $P\left(5,\frac{16}{3}\right)$ lies on $H$ . Verify that $PS=e\times PM$ , where $S$ is the focus $(5,0)$ and $M$ is the foot of the perpendicular from $P$ to the corresponding directrix.2 marks
An ellipse $E$ has its centre at the origin and its foci at $(\pm4,0)$ on the $x$ -axis. Its eccentricity is $\frac23$ .
Find the equation of $E$ in the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).