Parabola Tangent Problem

[Joe Mercurio]

Problem: Prove that the two tangents to a parabola at the ends of a chord through the focus are perpendicular to each other. The question asks me to use the reflection property of parabolas but I thought of solving it a little differently.

My attempted solution was to assume this was true and look for any inconsistencies. So Step 1 was to find the derivative at an arbitrary point on the curve. The negative inverse of this derivative should be the slope at the other end of the chord. So setting the derivative of the second point equal to the negative inverse of the derivative of the first point, I could work backwards and solve for Point 2. Then I'll have 2 points and can draw a line through them. If this is true, then the focus of the parabola should be a solution for this line and if the vertex is at the origin, the focus will be at the y intercept of that line.

Parabola: y=ax².
y'=2ax

I picked (\(\displaystyle x_{1}\),\(\displaystyle ax_{1}^{2}\)) as Point 1.
\(\displaystyle y_{1}^{'}\)=\(\displaystyle 2ax_{1}\).
\(\displaystyle y_{2}^{'}\) as the negative inverse is equal to \(\displaystyle \frac{-1}{2ax_{1}}\)

Solving for \(\displaystyle x_{2}\)
\(\displaystyle y_{2}^{'}\) = \(\displaystyle 2ax_{2}\)
\(\displaystyle x_{2}\) = \(\displaystyle \frac{-1}{4a^{2}x_{1}}\)
Point 2 = (\(\displaystyle \frac{-1}{4a^{2}x_{1}}\),\(\displaystyle \frac{1}{16a^{4}x_{1}^{2}}\))

Finding the slope of the line (m = \(\displaystyle \frac{y_{2}-y_{1}}{x_{2}-x_{1}}\))
Plugging in the Points found above I come up with a slope of:
m = \(\displaystyle \frac{1-16a^{5}x_{1}^{4}}{-4a^{2}x_{1}(1+4a^{2}x_{1}^{2})}\)

Now that I know m, and have one of the points I should be able to solve for B for the line y=mx+b. Sparing you the messy algebra i end up with:
b = \(\displaystyle \frac{4a^{3}x_{1}^{2}+1}{4a^{2}+16a^{4}x_{1}^{2}}\)

I was hoping b would equal out to be \(\displaystyle \frac{1}{4a}\) but that obviously isn't the case. Is there a flaw in my reasoning or did I just screw up the algebra?

 

[Joe Mercurio]

nevermind, I figured it out. Think I just needed to take a step back

The correct y coordinate for Point 2 is \(\displaystyle \frac{-a}{2ax_{1}}\) which allows the numerator to cancel with the denominator leaving \(\displaystyle \frac{1}{4a}\) behind. Thanks!

 

[soroban]

Hello, Joe!

I solved this problem years ago.
I'll show you my approach.

Prove that the two tangents to a parabola at the ends of a chord through the focus
are perpendicular to each other.

We are given the parabola: .\(\displaystyle x^2 \,=\,4py \quad\Rightarrow\quad y \:=\:\frac{1}{4p}x^2\)

The equation of the line through the focus \(\displaystyle (0,p)\) with slope \(\displaystyle m\) is: \(\displaystyle y \:=\:mx + p\)

They intersect when: .\(\displaystyle \frac{1}{4p}x^2 \:=\:mx + p \quad\Rightarrow\quad x^2 - 4pmx - 4p^2\:=\:0\)

Quadratic Formula: .\(\displaystyle x \;=\;\frac{4pm \pm \sqrt{16p^2m^2 + 16p^2}}{2} \;=\;2p(m \pm\sqrt{m^2+1})\)

We have: .\(\displaystyle \begin{Bmatrix}x_1 &=& 2p(m + \sqrt{m^2+1} \\ x_2 &=& 2p(m - \sqrt{m^2+1} \end{Bmatrix}\)


The slope of the tangent is: .\(\displaystyle y' \:=\:\frac{1}{2p}x\)

At \(\displaystyle x_1\!:\; m_1 \:=\:\frac{1}{2p}\cdot 2p(m + \sqrt{m^2+1}) \:=\:m + \sqrt{m^2+1}\)

At \(\displaystyle x_2\!:\;m_2 \:=\:\frac{1}{2p}\cdot2p(m-\sqrt{m^2+1}) \:=\:m - \sqrt{m^2+1}\)


\(\displaystyle m_1\cdot m_2 \;=\;(m+\sqrt{m^2+1})(m-\sqrt{m^2+1}) \:=\:m^2 - (m^2+1) \:=\:-1\)

. . Therefore: \(\displaystyle m_1 \perp m_2.\)

\(\displaystyle \text{Q.E.D.}\)

 

[Joe Mercurio]

Hi Soroban,

That was a neat approach, thanks for sharing. I'm also jealous of your LaTeX skills as well!

I concede the more elegant approach to you, but I'm curious if my method was sound as well? I'm worried I mistakenly arrived at the right conclusion