How to classify Conic Sections?

Advanced conic sections redux - (serious responses only please). What is the equation for this locus of points?

• Using polar coordinates (r, θ), an ellipse contains the points: (1, 1), (1, 1+2π/3), (1, 1+4π/3) What is the equation for the locus of points representing a major vertex of all possible such ellipses? Feel free to represent the equation in either polar or cartesian form. Prior conic section questions: http://answers.yahoo.com/question/index;_ylt=AgliPRAi69idOB8WMpUhXifsy6IX;_ylv=3?qid=20090817152417AAqayB4

• Answer:

  http://en.wikipedia.org/wiki/Ellipse#General_ellipse According to the above, the general equation of an ellipse could be Ax²+Bxy+Cy²+Dx+Ey=1 ..... (1), where B²-4AC<0 ==> The ellipse is defined by 5 points. With only 3 points given, there will be infinite many ellipses, which vertices will belong to a distinct areas of the xOy coordinate plane. The bounds of these areas will be: 1) The 3 lines defined by the 3 given points and 2) The curves where B²-4AC=0, which will separate the ellipses and the hyperbolas passing through the 3 given points and probably 3) The unit circle (not quite sure about that) After the clockwise rotation of the coordinate system for 1 radian first, so that the points are at 0, 2π/3, and 4π/3, their Cartesian coordinates are A(1, 0), B(-1/2,√(3)/2) and C(-1/2, -√(3)/2). Substituting the coordinates of the points A, B and C in (1) leads to the relations C=2 -A .....(2) D=1 -A ..... (3) E=B/2 ..... (4) The border curves between the vertices of the ellipses and the hyperbolas will be the ones on which the vertices of the parabolas through A, B and C lie, which leads to an additional relation for the 3 families of parabolas B=±2√(AC) ..... (5). So, my approach for answering this question will follow the same pattern as in your previous question, but with more complicated expressions, because of the position of the 3 given points. At first the locus of the vertices of the parabolas have to be found. In this case it is suitable to take the coefficient A as a parameter ==> The equation of the 3 families of parabolas will be Ax²±2√[A(2-A)]xy+(2-A)y²+ (1-A)x+√[A(2-A)]y=1 ..... (6) Let's start with the case when A≥C and B≥0. Obviously then1≤A≤2. Let α be the angle between the axis of a parabola and +x-axis ==> cot(2α)=(A-C)/B=(A-1)/√[A(2-A)] ==> cotα={0.5/√[A(2-A)]}*[A-1+√(-3A²+6A+1)…‡ The first derivative of (6) in this case is y'=-{2Ax+2√[A(2-A)]y+1-A}/ {2√[A(2-A)]x+(4-2A)Y-√[A(2-A)]}. At the vertex of a parabola y'=-cotα Edit: I am not sure if I'll finish that analytically, it's too complicated. If I have enough time I will sketch the region. I hope somebody else will answer this question. It's a nice question. If you extend the time for answering, may be I'll try my best to finish this, analytically or graphically, or both.