Do you know of any software I can use to graph this equation?

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The discussion centers around graphing an equation derived from the concept of conic sections, specifically when the directrix is a parabola represented by y=x^2. The user is attempting to establish a relationship between points on the curve and the focus, leading to a cubic equation that needs to be solved. They seek software recommendations for plotting this equation, ideally in a 2D format, and express interest in rearranging the equation to isolate y. Ultimately, they mention receiving assistance from another forum user who suggested using Mathematica for graphing. The conversation highlights the complexities of conic sections and the need for effective graphing tools.
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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)
YmiTxhg.png


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.
 
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