Multiplication On The Parabola

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Theorem

Let $\map f x = x^2$.

Let $A = \tuple {x_a, y_a}$ and $B = \tuple {x_b, y_b}$ be points on the curve $\map f x$ so that $x_a \le x_b$.

Then the line segment joining $A B$ will cross the $y$-axis at $-x_a x_b$.


Proof

Let $\map f x = x^2$.

Then:

$\map f {x_a} = x_a^2$

and:

$\map f {x_b} = x_b^2$

Then the slope of the line segment joining $A B$ will be:

\(\displaystyle m\) \(=\) \(\displaystyle \frac {x_a^2 - x_b^2} {x_b - x_a}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {x_a - x_b} \paren {x_b + x_a} } {x_b - x_a}\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle x_b + x_a\) cancelling


The result follows from taking the equation of the line defined by its slope and either point $A$ or $B$ to calculate the $y$-intercept.

Proceeding with point $A$:



\(\displaystyle x_a^2\) \(=\) \(\displaystyle \paren {x_b + x_a} x_a + b\)
\(\displaystyle x_a^2 - \paren {x_a + x_b} x_a\) \(=\) \(\displaystyle b\)
\(\displaystyle x_a^2 - x_a^2 - x_b x_a\) \(=\) \(\displaystyle b\)
\(\displaystyle -x_b x_a\) \(=\) \(\displaystyle b\)

$\blacksquare$