Deriving the Focus-Directrix Equation

From a Distance Definition to a Polar Equation

A conic can be described using a focus and a directrix: the distance from each point on the conic to the focus is a fixed multiple of its distance to the directrix. Polar coordinates are especially convenient when the focus is at the pole, because the distance to the focus is simply the radial distance.

The main work is expressing the distance to the directrix and handling its absolute value correctly. That sign matters: one choice gives the familiar polar equation for a branch, while a second choice can be necessary for a hyperbola.

In this tutorial, place the focus at the pole. A point with polar coordinates \((r,\theta)\), taking \(r\ge 0\), has Cartesian coordinates \(x=r\cos\theta\) and \(y=r\sin\theta\). Its distance to the focus is \(r\).

A Vertical Directrix

Take the vertical line \(x=d\), where \(d>0\). For a point \((x,y)\), the perpendicular distance to this line is \(|x-d|\). Thus the focus-directrix condition is

The absolute value expresses two possible locations: the point may lie on either side of the directrix. First consider the branch on the side \(x\le d\). There, \(|x-d|=d-x\), so

Collecting the terms involving \(r\) gives \(r(1+e\cos\theta)=ed\). When the denominator is nonzero, this gives the polar equation below. The condition \(x\le d\) is part of this derivation; the equation describes points on this side of the directrix.

To check the branch condition for points described by this equation, substitute \(x=r\cos\theta\):

Whenever \(1+e\cos\theta>0\), this is negative, as required. For \(e>1\), the equation applies only at angles where its denominator is positive and \(r\) is nonnegative.

On the other side, \(x\ge d\), the distance to the directrix is \(x-d\). The original condition then becomes \(r=e(r\cos\theta-d)\). Rearranging gives the other possible branch equation:

This equation requires \(e\cos\theta-1>0\), so that \(r\ge0\). Substitution gives \(x-d=d/(e\cos\theta-1)>0\), confirming that it describes points on the required side. If \(0<e\le1\), this side cannot contain a solution: \(r=e(r\cos\theta-d)\le e(r-d)<r\) when \(r\ge d\) and \(e\le1\), which contradicts the distance condition. When \(e>1\), the second branch can occur.

Horizontal Directrices and Reflections

The same distance argument works for a horizontal directrix. For the line \(y=d\), with \(d>0\), the perpendicular distance is \(|y-d|=|r\sin\theta-d|\). On the side \(y\le d\), it is \(d-y\). Equating \(r\) to \(e\) times that distance gives

For the other side, \(y\ge d\), the distance is \(y-d\). The corresponding equation is \(r=e(r\sin\theta-d)\), and hence

with \(e\sin\theta-1>0\) required for a nonnegative radius.

Moving a directrix to the negative side of the focus changes the sign in the corresponding expression. For example, with vertical directrix \(x=-d\), the side \(x\ge-d\) has distance \(x+d\). The condition \(r=e(x+d)\) yields \(r(1-e\cos\theta)=ed\), so the equation for that side is \(r=ed/(1-e\cos\theta)\). For horizontal directrix \(y=-d\), the side \(y\ge-d\) similarly gives \(r=ed/(1-e\sin\theta)\).

Why the Derivation Works

The derivation is an application of the definition, not a guess based on a memorized conic formula. For a vertical line, the perpendicular distance depends only on the horizontal coordinate. Since \(x=r\cos\theta\), the distance condition contains \(r\cos\theta\); since the focus is at the pole, the other distance is \(r\). Rearranging this one equation isolates \(r\).

Each step preserves the original geometric condition only when its sign assumption is respected. Removing the absolute value without specifying the side could silently discard a branch or introduce points that do not meet the distance definition.

Worked Examples

A Common Sign Error

A frequent mistake is to use the wrong coordinate sign when removing the absolute value. For directrix \(x=4\), a point with \(x\le4\) has distance \(4-x\), not \(4+x\). The incorrect calculation \(r=e(4+x)=e(4+r\cos\theta)\) would give \(r=4e/(1-e\cos\theta)\). That equation does not follow from the distance to \(x=4\).

For example, with \(e=\frac12\) and \(\theta=0\), the incorrect formula gives \(r=2/(1-\frac12)=4\), so \(x=4\). The actual focus-directrix condition at that point is \(r=\frac12|4-4|=0\), not \(4\). The correction is to start with \(r=e|r\cos\theta-4|\), then use \(4-r\cos\theta\) on the specified side. That gives the verified equation \(r=4/(2+\cos\theta)\).

Why the Distance Definition Matters

Deriving the equation from distances makes its signs and restrictions interpretable. The trigonometric term records the direction of the point relative to the directrix, while the eccentricity scales the directrix distance. Most importantly, retaining the side condition prevents a formula from being treated as though it automatically represents every point allowed by the absolute-value definition.

When deriving a polar conic equation, keep three items together: the directrix, the chosen side, and the resulting equation. For a horizontal directrix, replace the horizontal coordinate by \(r\sin\theta\); for a vertical one, use \(r\cos\theta\). Then check the result in the original distance condition.