From the Cartesian equation

Take P to be the origin. For a curve given by the equation F(x, y)=0, if the equation of the tangent line at R=(x₀, y₀) is written in the form

${\displaystyle \cos \alpha x+\sin \alpha y=p}$

then the vector (cos α, sin α) is parallel to the segment PX, and the length of PX, which is the distance from the tangent line to the origin, is p. So X is represented by the polar coordinates (p, α) and replacing (p, α) by (r, θ) produces a polar equation for the pedal curve.^[2]

Pedal curve (red) of an ellipse (black). Here a=2 and b=1 so the equation of the pedal curve is 4x²+y²=(x²+y²)²

For example,^[3] for the ellipse

${\displaystyle {\frac {x^{2}}{a^{2}}}+{\frac {y^{2}}{b^{2}}}=1}$

the tangent line at R=(x₀, y₀) is

${\displaystyle {\frac {x_{0}x}{a^{2}}}+{\frac {y_{0}y}{b^{2}}}=1}$

and writing this in the form given above requires that

${\displaystyle {\frac {x_{0}}{a^{2}}}={\frac {\cos \alpha }{p}},\,{\frac {y_{0}}{b^{2}}}={\frac {\sin \alpha }{p}}.}$

The equation for the ellipse can be used to eliminate x₀ and y₀ giving

${\displaystyle a^{2}\cos ^{2}\alpha +b^{2}\sin ^{2}\alpha =p^{2},\,}$

and converting to (r, θ) gives

${\displaystyle a^{2}\cos ^{2}\theta +b^{2}\sin ^{2}\theta =r^{2},\,}$

as the polar equation for the pedal. This is easily converted to a Cartesian equation as

${\displaystyle a^{2}x^{2}+b^{2}y^{2}=(x^{2}+y^{2})^{2}.\,}$

From the polar equation

For P the origin and C given in polar coordinates by r = f(θ). Let R=(r, θ) be a point on the curve and let X=(p, α) be the corresponding point on the pedal curve. Let ψ denote the angle between the tangent line and the radius vector, sometimes known as the polar tangential angle. It is given by

${\displaystyle r={\frac {dr}{d\theta }}\tan \psi .}$

Then

${\displaystyle p=r\sin \psi }$

and

${\displaystyle \alpha =\theta +\psi -{\frac {\pi }{2}}.}$

These equations may be used to produce an equation in p and α which, when translated to r and θ gives a polar equation for the pedal curve.^[4]

For example,^[5] let the curve be the circle given by r = a cos θ. Then

${\displaystyle a\cos \theta =-a\sin \theta \tan \psi }$

so

${\displaystyle \tan \psi =-\cot \theta ,\,\psi ={\frac {\pi }{2}}+\theta ,\alpha =2\theta .}$

Also

${\displaystyle p=r\sin \psi \ =r\cos \theta =a\cos ^{2}\theta =a\cos ^{2}{\alpha \over 2}.}$

So the polar equation of the pedal is

${\displaystyle r=a\cos ^{2}{\theta \over 2}.}$

From the pedal equation

The pedal equations of a curve and its pedal are closely related. If P is taken as the pedal point and the origin then it can be shown that the angle ψ between the curve and the radius vector at a point R is equal to the corresponding angle for the pedal curve at the point X. If p is the length of the perpendicular drawn from P to the tangent of the curve (i.e. PX) and q is the length of the corresponding perpendicular drawn from P to the tangent to the pedal, then by similar triangles

${\displaystyle {\frac {p}{r}}={\frac {q}{p}}.}$

It follows immediately that the if the pedal equation of the curve is f(p,r)=0 then the pedal equation for the pedal curve is^[6]

${\displaystyle f(r,{\frac {r^{2}}{p}})=0}$

From this all the positive and negative pedals can be computed easily if the pedal equation of the curve is known.

From parametric equations

Contrapedal of the same ellipse

Pedal of the evolute of the ellipse : same as the contrapedal of the original ellipse

Let ${\displaystyle {\vec {v}}=P-R}$ be the vector for R to P and write

${\displaystyle {\vec {v}}={\vec {v}}_{\parallel }+{\vec {v}}_{\perp }}$,

the tangential and normal components of ${\displaystyle {\vec {v}}}$ with respect to the curve. Then ${\displaystyle {\vec {v}}_{\parallel }}$ is the vector from R to X from which the position of X can be computed.

Specifically, if c is a parametrization of the curve then

${\displaystyle t\mapsto c(t)+{c'(t)\cdot (P-c(t)) \over |c'(t)|^{2}}c'(t)}$

parametrises the pedal curve (disregarding points where c' is zero or undefined).

For a parametrically defined curve, its pedal curve with pedal point (0;0) is defined as

${\displaystyle X[x,y]={\frac {(xy'-yx')y'}{x'^{2}+y'^{2}}}}$

${\displaystyle Y[x,y]={\frac {(yx'-xy')x'}{x'^{2}+y'^{2}}}.}$

The contrapedal curve is given by:

${\displaystyle t\mapsto P-{c'(t)\cdot (P-c(t)) \over |c'(t)|^{2}}c'(t)}$

With the same pedal point, the contrapedal curve is the pedal curve of the evolute of the given curve.

Example

Limaçon — pedal curve of a circle

When C is a circle the above discussion shows that the following definitions of a limaçon are equivalent:

• It is the pedal of a circle.
• It is the envelope of circles whose diameters have one endpoint on a fixed point and another endpoint which follow a circle.
• It is the envelope of circles through a fixed point whose centers follow a circle.
• It is the roulette formed by a circle rolling around a circle with the same radius.

We also have shown that the catacaustic of a circle is the evolute of a limaçon.