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Thread: Prove a Parabola Property

  1. #1
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    Prove a Parabola Property

    Problem 31 of chapter 12, section 1 of Calculus with Analytic Geometry by Purcell & Varberg, fifth edition:

    Prove that the tangents to a parabola at the extremities of any focal cord are perpendicular to each other.

    I solved this problem geometrically, but I first tried it algebraically and got an unexpected result:

    I started with a concave up parabola with p as the distance from the vertex to the focus: $$x^2 = 4py$$

    With `m` for the slope of the focal cord: $$y = mx + p$$

    To determine the points on the parabola for a focal cord I substitute y in the parabola's formula: $$x^2 = 4p(mx + p) = 4pmx + 4p^2 \longrightarrow x^2 - 4pmx - 4p^2 = 0 \longrightarrow x = 2pm 2p\sqrt{m^2 + p^2}$$

    This gives the points $$x_1 = 2pm - 2p\sqrt{m^2 + p^2}, y_1 = 2pm^2 + p^3 - 2pm\sqrt{m^2 + p^2}$$
    $$x_2 = 2pm + 2p\sqrt{m^2 + p^2}, y_2 = 2pm^2 + p^3 + 2pm\sqrt{m^2 + p^2}$$

    From the slope of the tangent to the parabola: $$\frac{dy}{dx} = \frac{x}{2p}, m_1 = m - \sqrt{m^2 + p^2}, m_2 = m + \sqrt{m^2 + p^2}$$

    From the proposition: $$m_1 = -\frac{1}{m_2} \longrightarrow (m - \sqrt{m^2 + p^2})(m + \sqrt{m^2 + p^2}) = -1 \longrightarrow p^2 = 1$$

    This would suggest that the proposition is true only when the absolute value of p is 1. Where did I go wrong?
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    Re: Prove a Parabola Property

    Quote Originally Posted by Zexuo View Post
    Problem 31 of chapter 12, section 1 of Calculus with Analytic Geometry by Purcell & Varberg, fifth edition:

    Prove that the tangents to a parabola at the extremities of any focal cord are perpendicular to each other.

    I solved this problem geometrically, but I first tried it algebraically and got an unexpected result:

    I started with a concave up parabola with p as the distance from the vertex to the focus: $$x^2 = 4py$$

    With `m` for the slope of the focal cord: $$y = mx + p$$

    To determine the points on the parabola for a focal cord I substitute y in the parabola's formula: $$x^2 = 4p(mx + p) = 4pmx + 4p^2 \longrightarrow x^2 - 4pmx - 4p^2 = 0 \longrightarrow x = 2pm 2p\sqrt{m^2 + p^2}$$
    Shouldn't that last $\sqrt{m^2 + p^2}$ be $\sqrt{m^2 + 1}$?
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    Re: Prove a Parabola Property

    Quote Originally Posted by Walagaster View Post
    Shouldn't that last $\sqrt{m^2 + p^2}$ be $\sqrt{m^2 + 1}$?
    I omitted the initial quadratic formula in the OP: $$\frac{4pm \sqrt{16p^2m^2 + 16p^4}}{2}$$

    A p^2 term remains under the radical after pulling 16p^2 out of it.
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    Re: Prove a Parabola Property

    But the $16p^4$ is what is wrong. It should be $16p^2$.
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    Re: Prove a Parabola Property

    Right you are! Thanks for the catch.
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