# Multiplication On The Parabola

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## Theorem

Let $f \left({x}\right) = x^2$.

Let $A = \left({x_a, y_a}\right)$ and $B = \left({x_b, y_b}\right)$ be points on the curve $f \left({x}\right)$ so that $x_a \leq x_b$.

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

## Proof

Let $f \left({x}\right) = x^2$.

Then:

- $f \left( {x_a} \right) = x_a^2$

and:

- $f \left( {B_x} \right) = x_b^2$

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

\(\displaystyle m\) | \(=\) | \(\displaystyle \frac{x_a^2 - x_b^2} {x_b - x_a}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac{\left( {x_a - x_b} \right) \left( {x_b + x_a} \right)} {x_b - x_a}\) | 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 $A$:

\(\displaystyle x_a^2\) | \(=\) | \(\displaystyle \left( {x_b + x_a}\right) x_a + b\) | |||||||||||

\(\displaystyle x_a^2 - \left( {x_a + x_b} \right) x_a\) | \(=\) | \(\displaystyle b\) | |||||||||||

\(\displaystyle x_a^2 - x_a^2 - x_b x_a\) | \(=\) | \(\displaystyle b\) | |||||||||||

\(\displaystyle - x_b x_a\) | \(=\) | \(\displaystyle b\) |

$\blacksquare$