Let $\alpha(t) = (f(t), g(t)), t \in \mathbb{R}$ be a regular curve and $P(x_0, y_0)$ be a fixed point in the cartesian plane. The set of points $X$ such that the line $PX$ is perpendicular to the tangent line at $P$ is called the pedal curve of $\alpha$ in relation to $P$. Obtain a parametrization of $\alpha$ in relation to $P$.

How I did it:

At any given $t$, the line going through $(f(t), g(t))$ with directional vector $(f'(t), g'(t))$ is given by $y = g(t) + \cfrac{\left(x - f(t)\right)}{f'(t)} \cdot g'(t)$

Similarly, the line going through $P(x_0, y_0)$ perpendicular to the line above has the directional vector $(-g'(t), f'(t))$, so it's equation is given by:

$y = y_0 + \cfrac{\left(x_0 - x\right)}{g'(t)} \cdot f'(t)$

We can obtain the intersection by setting them equal to each other, and so I get this incredibly ugly expression:

If we let $w = (x(t), y(t))$ be the pedal curve asked, we get:

$x(t) = \cfrac{y_0 + x_0 \cdot \cfrac{f'(t)}{g'(t)} +\cfrac{f(t)}{f'(t)} \cdot g'(t) - g(t)}{\cfrac{f'(t)}{g'(t)} + \cfrac{g'(t)}{f'(t)}}$

$y(t) = g(t) + \cfrac{\left(x - f(t)\right)}{f'(t)}\cdot g'(t)$

So the exercise is done. I tested this "formula" for a few cases and it does work, so I'm pretty sure I haven't made any mistakes. But I'm wondering if there is an easier way that gives a more elegant answer ('cause all that junk is really not pretty).


The pedal curve is the projection of $p-\alpha(t)$ on $\alpha’(t)$, i.e., $$\alpha(t)+\langle\alpha’(t),p-\alpha(t)\rangle\frac{\alpha’(t)}{\|\alpha’(t)\|^2}.$$

Moral: Avoid coordinates whenever possible.

  • $\begingroup$ Got it. Thanks! $\endgroup$ Jan 4 '18 at 21:14

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