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We know that the general equation of a conic is given by $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$ We also know that for this conic to be a parabola, the essential conditions are that $$\Delta = \begin{vmatrix} a&h&g\\ h&b&f\\ g&f&c\\ \end{vmatrix} \ne 0$$ and $$h^2 = ab$$ I am interested to find the proof of this.

Also we have been told that to find the center of the conic, perform partial differentiation on the given equation of the conic with respect to x and y separately. Then upon solving the two equations we can find the center. I have checked this very useful and convenient method, but I didn't find the proof of this concept in any standard book or the web. Can anyone please help me with the proof?

MathsLearner
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  • I'm not sure, but I think the determinant condition ensures that the geometric object defined by the equation is not degenerate. To obtain the second equation, consider $y=ax^2$ and see what Euclidean transformations (scaling by a real number, translations, rotations and reflections) do to it to find the general form of a hyperbola. – stressed out Sep 14 '18 at 09:17
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    For the first question: $h^2-ab$ is invariant under rotation, and for an axis-aligned one $h=0$. This means that in addition one of the coefficients of $x^2$ or $y^2$ is zero, and we have a parabola. – Jan-Magnus Økland Sep 14 '18 at 09:19
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    FYI: a parabola has no center. – Intelligenti pauca Sep 14 '18 at 13:48
  • For your second question, see the related question from the handy list at right. (Please try to stick to one question at a time, though.) – amd Sep 14 '18 at 19:45

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