What software should I use to graph this? / How do I get rearrange this equation so that it is in terms of y?

Question

I thought I'd just quickly tell you guys why I want to graph this equation before giving it you.

We're studying conic sections at the moment, and I started wondering what would happen if I let the directrix be a parabola. I looked at the instance where the directrix is y=x^2. I let the focus, F, be (x_F,y_F). Let P=(x,y) be a point on the curve that I am trying to find. I was wondering what the equation of this curve would be if I said that the distance between P and F and the perpendicular distance between P and the directrix has to be the same for any point P. ie. if we let O be the point where P meets the directrix at right angles, then OP=PF. I shall let O=(x_1,y_1)

The gradient of the directrix at point O has to 2(x_1). Thus the gradient of OP has to be (-1/(2(x_1))). Thus, we can represent O as (x+t, y-(t/(2(x_1))), where t is an unknown variable. Since x_1=x+t, we can further represent O as (x+t, y-(t/(2(x+t))). Substituting these values for x_1 and y_1 back into our equation for the directrix, we get:

y-(t/(2(x+t))=(x+t)^2

I let u=x+t, now: y-(u-x)/(2u)=u^2 => 2uy-(u-x)=u^3 => u^3+(1-2y)u-x=0

I then solve this cubic using Cardano's method (method in the link below) http://www.math.cornell.edu/~henderson/courses/M403-S03/cubics.htm

I get u=(((x/2)+((x/2)^2+((1-2y)/3)^3)^0.5)^(1/3))+(((x/2)-((x/2)^2+((1-2y)/3)^3)^0.5)^(1/3)) thus t=(((x/2)+((x/2)^2+((1-2y)/3)^3)^0.5)^(1/3))+(((x/2)-((x/2)^2+((1-2y)/3)^3)^0.5)^(1/3))-x

I want to show that OP=PF. (PF)^2= (x_F-x)^2 + (y_F-y)^2, and (OP)^2 = (y-y_1)^2 + (x-x_1)^2 = (t/(2(x+t)))^2 + t^2

Thus, replacing t in terms of x and y we get: (((((x/2)+((x/2)^2+((1-2*y)/3)^3)^0.5)^(1/3))+(((x/2)-((x/2)^2+((1-2*y)/3)^3)^0.5)^(1/3))-x)^2)*((((x/2)+((x/2)^2+((1-2*y)/3)^3)^0.5)^(1/3))+(((x/2)-((x/2)^2+((1-2*y)/3)^3)^0.5)^(1/3))+1)/(4*((((x/2)+((x/2) ^2+(1-2*y)^3)^0.5)^(1/3))+(((x/2)-((x/2)^2+(1-2*y)^3)^0.5)^(1/3)))^2)-(x_F-x)^2-(y_F-y)^2 = 0

(easier to see equation)

This is the function that I want to plot. I'm just interested to see what it'd look like. Of course I'd have to replace x_F and y_F with set values. I just want to know if any of you know of software which can plot this equation. I don't want to plot a 3-D graph though, unless I am able to set the value of z to 0.

OR

If you guys know of a way to rearrange this equation so that y becomes the subject, then I can simply use any graphing software to plot the curve.

Answer 1

I will use somewhat different notation, for clarity. Let $F = (p,q)$ be a "focus," and for each point $P(t) = (t,t^2)$ on the parabola $y = x^2$, suppose we are interested in the locus of points $L(p,q,t)$ such that $PL = LF$ where $PL$ is normal to the parabola. This is a complicated curve, but it can be parametrized as $$L(p,q,t) = \left(\frac{p^2 t+q^2 t-2 q t^3-q t+t^5}{2 p t-q-t^2},\frac{-p^2+4 p t^3+2 pt-q^2-3 t^4-t^2}{2 \left(2 p t-q-t^2\right)}\right).$$ Note that this requires $t^2-2pt + q \ne 0$.

We can use Mathematica to draw an interactive plot of this locus:

F[p_,q_,t_] := {(p^2 t - q t + q^2 t - 2 q t^3 + t^5)/(-q + 2 p t - t^2),
       (p^2 + q^2 - 2 p t + t^2 - 4 p t^3 + 3 t^4)/(2 (q - 2 p t + t^2))}
Manipulate[Show[ParametricPlot[{F[p, q, t], {t, t^2}}, {t, -3, 3}, 
    PlotRange -> {{-5, 5}, {0, 10}}, Exclusions -> {t^2 == 2 p t - q}],
    Graphics[{Disk[{p, q}, .05], Line[{{m, m^2}, F[p, q, m], {p, q}}]}]],
    {{p, -1}, -3, 3}, {{q, 2}, -3, 3}, {{m, 0}, -3, 3}]

which gives, for instance, the following:

Here is a picture for varying $p$: