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How to define ShapeDrawables programmatically?

How many points are needed to uniquely define a parabola?

  • and the other conics? Is there only one parabola passing through the vertices of a given triangle? ====================================== some thoughts on the question: - 1 point as focus/vertex and 1 line as directrix define a parabola. So i don't think that just one vertex and a slope can define one because you can draw as many lines as you want with the given slope and have differents parabolas - The equation of a general parabola is: Ax² +/- 2sqrt(|AC|) xy + Cy²+ Dx + Ey + F = 0 so it seems to me that are needed 5 points to define the curve, but i'm not sure if there's a transformation that reduces it's quantity by 4 or 3 ========================================== I see it now, if we fix A=1 we can have: x² + 2sqrt(C)xy + Cy² + Dx + Ey + F = 0 we have a quadratic equation for sqrt(C) and because its not linear we can't say 4 points define one parabola. But most of time 4 points will do it

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    EDIT: I miscounted at first, as pointed out by . Correct answer below. Four points generally suffice, though usually there could be two parabolas through the four points rather than just one. There's some confusion here in high-school math, where it is often tacitly assumed that the directrix of the parabola is vertical, or in other words that the parabola is defined by [math]ax^2+bx+c=0[/math]. Students and teachers sometimes conflate "parabola" with "quadratic polynomial in one variable", in which case three points would indeed suffice to pin it down. There's a unique circle through any three points (again, allowing for degeneracy).

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As you've noted in the details to the question, a parabola has an equation of the form            [math]Ax^2\pm 2\sqrt{AC}\, xy + Cy^2+ Dx + Ey + F = 0[/math] That's a special case of the generic equation for a conic section            [math]Ax^2+Bxy+Cy^2+Dx+Ey+F=0[/math] For a parabola, the discriminant  [math]B^2-4AC[/math]  has to be 0; for an ellipse it's negative; and for a hyperbola, it's positive. Although this generic equation has 6 variables, A through F, that doesn't imply that there are 6 degrees of freedom when you specify a conic section. There are only 5 since scaling all the variables up or down by the same factor doesn't change the curve.  So, 5 generic points determine a conic section, something that can be shown geometrically as well. Since there's that extra equation for a parabola, [math]B^2=4AC,[/math] there's one fewer degrees of freedom in the specification of a parabola.  Therefore, there are 4 degrees of freedom to specify a parabola. How can you use these 4 degrees of freedom to specify a particular parabola?  Use two of them to specify a point to be a focus. Two degrees of freedom are used to specify that point.  Use two more to determine the line which is the directrix. You can do  that with one more point if you take the directrix to be the line through that point perpendicular to the line joining the two points. Thus, you can use two points to define a parabola.

David Joyce

Well, two things make a parabola. It has A vertex, which is basically its starting point, and once you have this you need A slope. A slope is the direction and speed that you go away from the starting point (up, down, fast, slow). Anything that defines an exact spot for a vertex and the slope can graph a variety of lines, namely stuff like y= x, y= x^2, y= x^4, anything with slopes that are easy to graph (odd exponents tend to go up and down a lot so you would have to be more specific than one that keeps going up). Thought experiment time. Let's say you had a flashlight. If you shined it and paid attention only to the outer rims of the light, then you would have a type of 3-d parabola, right? When you step forward, you completely change where the outer rim of the light touches. Once you approach a wall and get right next to it the light will barely shine on anything, right? Once you are a hundred feet away from the wall, the whole wall will be illuminated. This is a good way to think about how to graph parabolas (you being the vertex, or distance from the wall, and the light being the slope, continually growing larger and being very unique to your flashlight. If someone had a pocket light and you had a spotlight, standing side by side, regardless of where you are the wall would be illuminated in WAY different amounts.) Ok end of thought experiment. Tl;dr Flashlights are parabolas. Picture getting closer to a wall with different flashlights.

Kyle Certiorari

Two. Given A and B, draw a line through B, perpendicular to the vector AB. Your parabole is the locus of all points for which the distance to the line equals the distance to A

Gerwin Dox

I think you need 3 points to uniquely identify a parabola, 2 points to define the directrix line and the other one the focus, therefore you get the equation as. If P(x,y) is any point on the parabola. Distance of the line from the point P = distance of the focus from point P.

Deepak Kumar

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