Equation of locus of moving point

Question

How to find the equation of locus of a moving point such that its perpendicular distance from a function $f(x)$ is always $g(x)$?

Is $f$ differentiable on its domain? – Mikasa Sep 26 '13 at 08:45 — Mikasa, Sep 26 '13 at 08:45
yeah $f$ differentiable on its domain – Tom Lynd Sep 26 '13 at 09:00 — Tom Lynd, Sep 26 '13 at 09:00

score 5 · Accepted Answer · answered Sep 26 '13 at 08:57

5

Consider the plot below:

enter image description here

So, I assumed that $f$ is differentiable on its domain. We know the equation of the line $l$ as follows:

$$l:~~-f'(x_0)x+y+(f'(x_0)x_0-y_0)=A_1x+y+B_1=0$$

Since the distance $d$ between the points $(X,Y)$ and the straight line $l$ is given by $$d=\frac{|A_1X+Y+B_1|}{\sqrt{A_1^2+1}}$$ so it seems that we should consider $$d=g(x)$$

answered Sep 26 '13 at 08:57

Mikasa

67,374

Great "handiwork" and text to follow! +1 – amWhy Sep 27 '13 at 00:50
@amWhy: Thanks dear. :-) – Mikasa Sep 27 '13 at 05:17

Equation of locus of moving point

1 Answers1