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Thread: What is the maximum number of intersects on a coplaner parabola and a hyperbola?

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    Level 1 - Newbie divardin's Avatar
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    What is the maximum number of intersects on a coplaner parabola and a hyperbola?

    What is the maximum number of intersects on a coplaner parabola and a hyperbola?

    I know the relevant answer is 4, however, someone asked if it was possible that the parabola took the same curve path of one of the hyperbolas. Is it possible that in that case, the maximum number could be infinite?

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  3. #2
    Level 1 - Newbie noel's Avatar
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    The correct answer is 4 indeed.
    The Cartesian coordinates of common points of two 2nd degree curves are the solutions of the system of two 2nd degree equations with 2 unknowns. The elimination of one of them leads to a single algebraic equation with 1 unknown of at most 4th degree that may have 0, 1, 2, 3, 4 or infinitely many (the latter case possible if it is identity) real roots.

    The equations of the parabola and the hyperbola are structurally different (they have different discriminants - follow the link in Sources below for details), what leads to different geometrical properties of both curves, for example:
    1) Parabola has a single branch, hyperbola has two;
    2) Parabola has no asymptotes, hyperbola has two;
    3) Parabola's eccentricity is 1, hyperbola's - greater than 1 (see an animation in the article, 'Parameters' Section), etc.
    That's why it is impossible for a parabola to be identical with a branch of hyperbola, so both curves can not have more than 4 common points.

    Here is an example:
    parabola y² = x + 11 /"c"-shaped/,
    hyperbola x² - y² = 1 /") ("-shaped/,
    Common points (-3, -?8), (-3, ?8), (4, -?15) and (4, ?15)

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