# Thread: Prove a Parabola Property

1. ## 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?

2. ## Re: Prove a Parabola Property

Originally Posted by Zexuo
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 $\sqrt{m^2 + 1}$?

3. ## Re: Prove a Parabola Property

Originally Posted by Walagaster
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.

4. ## Re: Prove a Parabola Property

But the $16p^4$ is what is wrong. It should be $16p^2$.

5. ## Re: Prove a Parabola Property

Right you are! Thanks for the catch.