FP1 4.2 - Foci, directrices and eccentricity | Edexcel A-Level Further Mathematics

Describe a conic using the ratio of distances from a focus and a directrix. You will connect eccentricity with the Cartesian parameters, find foci and directrices, and use shared geometric information to determine unknown conics.

One distance ratio creates three kinds of conic

An ellipse has equation $x^2/25+y^2/16=1$ . A hyperbola has the same foci and the reciprocal eccentricity. Can you find its two parameters without starting with simultaneous equations? The foci provide the useful shortcut; we will find $a=9/5,\ b=12/5$ .

You already know

Recognise the standard Cartesian and parametric conics from FP1 4.1. The distance between two points uses the square root of the sum of squared coordinate differences. A distance to a line is perpendicular: to $x=d$ it is $|x-d|$ , and to $x+y=d$ it is $|x+y-d|/\sqrt2$ .

Fix a point $F$ , called the focus, and a line $l$ , called the directrix, with $F$ not on $l$ . For a point $P$ , let $D$ be the foot of the perpendicular from $P$ to $l$ . The focus-directrix property is

PF=e\,PD

The positive constant $e$ is the eccentricity. The ratio uses ordinary non-negative distances, not signed horizontal displacements.

Eccentricity and conic type

For the non-degenerate focus-directrix conics,

\frac{PF}{PD}=e,\qquad \begin{cases} 0<e<1&\text{ellipse}\\ e=1&\text{parabola}\\ e>1&\text{hyperbola} \end{cases}

The printed table gives these types and their parameter relations. A circle is the limiting ellipse with $e=0$ ; it has no finite directrix in this description.

In the formula booklet.

An ellipse or hyperbola has two corresponding focus-directrix pairs. Use one pair consistently: the positive focus goes with the positive directrix in the horizontal standard form. Do not measure to whichever focus or line happens to be nearest.

Key idea

Fix one focus and its corresponding directrix. The conic consists of points with the same non-negative distance ratio $PF/PD=e$ , using perpendicular distance for $PD$ .

Check the chosen focus-directrix pair geometrically before calculating either distance.

[DIAGRAM: asset_name: Perpendicular distance to an ellipse directrix; asset_slug: 9fm0-fp171-focus-distance; caption: Eccentricity compares a focus distance with a perpendicular line distance; description: An equal-scale graph shows the ellipse x squared over 25 plus y squared over 9 equals 1, its right focus F at (4,0), and the vertical directrix x equals 25 over 4. The point P is (3,12 over 5) and its perpendicular foot D is (25 over 4,12 over 5). A blue segment PF and a sage horizontal segment PD are shown with a right-angle marker at D. A separate side panel gives PF equals 13 over 5, PD equals 13 over 4, and PF over PD equals 4 over 5. All distance labels are outside the curve.]
Eccentricity compares a focus distance with a perpendicular line distance

Watch out

Wrong: take $e$ to be the distance between the foci. Right: eccentricity is a dimensionless ratio. Enlarging a conic changes its lengths but not its eccentricity.

Use the ratio definition directly.

Equal distances give the parabola

For focus $F=(a,0)$ and directrix $x=-a$ , the property $PF=PD$ gives

\begin{aligned} \sqrt{(x-a)^2+y^2}&=|x+a| && \text{two distances}\\ (x-a)^2+y^2&=(x+a)^2 && \text{square non-negative quantities}\\ y^2&=4ax && \text{expand and cancel} \end{aligned}

Conversely, this equation gives the squared distance equality, and non-negative square roots recover the original distance equality. No extra branch is introduced.

Parabola geometry

For $y^2=4ax$ , where $a>0$ , the focus is $(a,0)$ , the directrix is $x=-a$ and $e=1$ . The vertex is halfway between the focus and directrix along the symmetry axis.

In the formula booklet.

The focus and directrix are on opposite sides of the vertex.

Worked example 1: an equation from equal distances

Question. Find the locus of points equidistant from $F=(2,0)$ and the line $x=-2$ , identifying the curve.

Plan. Set the point distance equal to the perpendicular line distance and square.

\begin{aligned} \sqrt{(x-2)^2+y^2}&=|x+2| && e=1\\ (x-2)^2+y^2&=(x+2)^2 && \text{both sides non-negative}\\ y^2&=8x && \text{cancel the matching terms} \end{aligned}

Why this step? The distance to the vertical line is $|x+2|$ , independent of $y$ . Using a sloping distance to a point on the line would describe a different locus.

Answer. The parabola $y^2=8x$ , with vertex $(0,0)$ , focus $(2,0)$ , directrix $x=-2$ , eccentricity $1$ .

Check. At $(2,4)$ , the focus distance is $4$ and the line distance is $4$ .

Change only the common focus/directrix scale from $2$ to $3$ .

Worked example 2: a wider distance scale

Question. Find the locus equidistant from $F=(3,0)$ and $x=-3$ .

Plan. Use the same equality with the new positions.

\begin{aligned} (x-3)^2+y^2&=(x+3)^2 && \text{squared distance equality}\\ y^2&=12x && \text{expand and subtract} \end{aligned}

Why this step? The eccentricity remains $1$ because the distances remain equal; changing the scale does not change the type.

Answer. $y^2=12x$ , a parabola with focus $(3,0)$ and directrix $x=-3$ .

Check. Its point $(3,6)$ is $6$ units from both the focus and the line.

The printed geometry also works after translating or interchanging coordinates. For example, $(y-k)^2=4a(x-h)$ has focus $(h+a,k)$ and directrix $x=h-a$ ; these translated forms follow by moving the whole picture.

Watch out

Wrong: the focus of $y^2=12x$ is $(12,0)$ . Right: compare with $4ax$ ; the focus is $(3,0)$ .

Change the distance scale once more.

Ellipse geometry from its semiaxes

For a horizontal major axis, take $a>b>0$ . The ellipse relation is $b^2=a^2(1-e^2)$ , so its eccentricity is the positive square root $e=\sqrt{1-b^2/a^2}$ .

To see the focus-directrix property, choose $F=(ae,0)$ and $x=a/e$ . The equation $PF^2=e^2PD^2$ becomes

\begin{aligned} (x-ae)^2+y^2&=e^2(x-a/e)^2\\ (1-e^2)x^2+y^2&=a^2(1-e^2)\\ \frac{x^2}{a^2}+\frac{y^2}{a^2(1-e^2)}&=1 \end{aligned}

Thus the denominator of $y^2$ is exactly $b^2$ . Since distances and $e$ are non-negative, the squared relation is equivalent to $PF=ePD$ .

Horizontal ellipse geometry

For $x^2/a^2+y^2/b^2=1$ , with $a>b>0$ ,

b^2=a^2(1-e^2),\quad 0<e<1,\qquad F=(\pm ae,0),\qquad x=\pm\frac ae

The two foci lie inside the ellipse; the two directrices lie outside it. If the major axis is vertical, interchange the coordinate roles before applying the formulas.

In the formula booklet.

Work with $e^2$ until the final positive square root. The symbol $e$ denotes a ratio, so it has no negative alternative.

Worked example 3: a rational eccentricity

Question. Find the eccentricity, foci and directrices of $x^2/25+y^2/16=1$ .

Plan. Read $a=5,b=4$ , calculate $e$ , then use $ae$ and $a/e$ for different purposes.

\begin{aligned} e^2&=1-\frac{16}{25}=\frac9{25} && \text{ellipse relation}\\ e&=\frac35,\qquad ae=3,\qquad \frac ae=\frac{25}3 && \text{positive ratio} \end{aligned}

Why this step? Multiplication gives the focus position, while division gives the directrix position. They are not interchangeable.

Answer. $e=3/5$ ; foci $(\pm3,0)$ ; directrices $x=\pm25/3$ .

Check. $3<5<25/3$ : each positive focus is inside its positive vertex and the corresponding directrix is beyond it.

Change only the vertical semiaxis from $4$ to $3$ .

Worked example 4: the foci move outwards

Question. Find the eccentricity, foci and directrices of $x^2/25+y^2/9=1$ .

Plan. Use the new $b^2$ , keeping $a=5$ .

\begin{aligned} e^2&=1-\frac9{25}=\frac{16}{25} && \text{changed vertical scale}\\ e&=\frac45,\qquad ae=4,\qquad \frac ae=\frac{25}4 && \text{geometry} \end{aligned}

Why this step? Reducing the vertical semiaxis with fixed horizontal semiaxis increases $e$ , so the foci move towards the vertices.

Answer. $e=4/5$ ; foci $(\pm4,0)$ ; directrices $x=\pm25/4$ .

Check. For $P=(3,12/5)$ , $PF=13/5$ and $PD=13/4$ for the positive pair; the ratio is $4/5$ , as in the diagram.

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FP1 4.2 - Foci, directrices and eccentricity

One distance ratio creates three kinds of conic

Equal distances give the parabola

Ellipse geometry from its semiaxes

Continue this lesson.

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