# A Parabola Property

• Dec 28th 2010, 04:08 AM
Soroban
A Parabola Property

Triangle $\displaystyle PQR$ is inscribed in parabola $\displaystyle y \,=\,ax^2.$

Tangents at $\displaystyle P,Q,R$ intersect at $\displaystyle P'\!,Q'\!,R',$ forming triangle $\displaystyle P'Q'R'.$

Show that the areas of the two triangles are in the ratio $\displaystyle 2:1.$

• Dec 28th 2010, 11:16 AM
Opalg
Quote:

Originally Posted by Soroban
Triangle $\displaystyle PQR$ is inscribed in parabola $\displaystyle y \,=\,ax^2.$

Tangents at $\displaystyle P,Q,R$ intersect at $\displaystyle P'\!,Q'\!,R',$ forming triangle $\displaystyle P'Q'R'.$

Show that the areas of the two triangles are in the ratio $\displaystyle 2:1.$

I'll do this for the parabola $\displaystyle y^2=4ax$ since I'm more familiar with that. But the proof applies to any parabola.

Take the three points to be $\displaystyle (ar^2,2ar),\ (as^2,2as,\ (at^2,2at).$ The area of the triangle PQR is half the absolute value of $\displaystyle \begin{vmatrix}ar^2&2ar&1\\ as^2&2as&1\\ at^2&2at&1\end{vmatrix}.$ Using elementary row operations and expanding down the right-hand column, this comes out to be

$\displaystyle \begin{vmatrix}ar^2&2ar&1\\ a(s^2-r^2)&2a(s-r)&0\\ a(t^2-r^2)&2a(-r)t&0\end{vmatrix} = 2a^2(s-r)(t-r)\begin{vmatrix}s+r&1\\t+r&1\end{vmatrix} = 2a^2(s-r)(t-r)(s-t).$

The tangent at R is $\displaystyle x-ty+at^2 = 0$. It meets the tangent at S at the point $\displaystyle P' = (ast,a(s+t))$. Similarly, $\displaystyle Q' = (atr,a(t+r))$ and $\displaystyle R' = (ars,a(r+s)).$ The area of the triangle P'Q'R' is half the absolute value of

$\displaystyle \begin{vmatrix}ast&a(s+t)&1\\ atr&a(t+r)&1\\ ars&a(r+s)&1\end{vmatrix} = \begin{vmatrix}ast&a(s+t)&1\\ at(r-s)&a(r-s)&0\\ as(r-t)&a(r-t)&0\end{vmatrix} = a^2(r-s)(r-t)\begin{vmatrix}t&1\\s&1\end{vmatrix} = a^2(r-s)(r-t)(t-s).$

Thus area(PQR) = 2area(P'Q'R').
• Dec 28th 2010, 11:20 AM
red_dog
Let $\displaystyle P(x_1,ax_1^2), \ Q(x_2,ax_2^2), \ Q(x_3,ax_3^2)$

The slopes of the tangents at $\displaystyle P, \ Q, \ R$ are:

$\displaystyle m_1=2ax_1, \ m_2=2ax_2, \ m_3=2ax_3$

The equations of the tangents are:

$\displaystyle y-ax_1^2=2ax_1(x-x_1)$

$\displaystyle y-ax_2^2=2ax_1(x-x_2)$

$\displaystyle y-ax_3^2=2ax_1(x-x_3)$

Solving the systems formed by two of three equations we find the coordinates of the points $\displaystyle P', \ Q', \ R'$:

$\displaystyle P'\left(\dfrac{x_2+x_3}{2},ax_2x_3\right), \ Q'\left(\dfrac{x_1+x_3}{2},ax_1x_3\right), \ R'\left(\dfrac{x_1+x_2}{2},ax_1x_2\right)$

The area of the triangle $\displaystyle P'Q'R'$ is $\displaystyle A_{P'Q'R'}=\dfrac{1}{2}|\Delta '|$

where $\displaystyle \Delta '=\begin{vmatrix}\dfrac{x_1+x_2}{2} & ax_1x_2 & 1\\ \dfrac{x_2+x_3}{2} & ax_2x_3 & 1\\ \dfrac{x_1+x_3}{2} & ax_1x_3 & 1\end{vmatrix}=\dfrac{a}{2}\begin{vmatrix}x_1+x_2 & x_1x_2 & 1\\ x_2+x_3 & x_2x_3 & 1\\ x_1+x_3 & x_1x_3 & 1\end{vmatrix}=\dfrac{a}{2}(x_1-x_2)(x_3-x_2)(x_3-x_1)$

Then $\displaystyle A_{P'Q'R'}=\dfrac{1}{4}|a(x_1-x_2)(x_2-x_3)(x_3-x_1)|$

But $\displaystyle A_{PQR}=\dfrac{1}{2}|\Delta|$

where $\displaystyle \Delta=\begin{vmatrix}x_1 & ax_1^2 & 1\\ x_2 & ax_2^2 & 1\\ x_3 & ax_3^2 & 1\end{vmatrix}=a\begin{vmatrix}1 & 1 & 1\\x_1 & x_2 & x_3\\x_1^2 & x_2^2 & x_3^2\end{vmatrix}=a(x_3-x_2)(x_3-x_1)(x_2-x_1)$

Then $\displaystyle A_{PQR}=\dfrac{1}{2}|a(x_1-x_2)(x_2-x_3)(x_3-x_1)|$

and $\displaystyle A_{P'Q'R'}=\dfrac{1}{2}A_{PQR}$