Multiplication using Parabola

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Theorem

Multiplication-using-Parabola.png

Let the parabola $P$ defined as $y = x^2$ be plotted on the Cartesian plane.


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


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


Thus $P$ can be used as a nomogram to calculate the product of two numbers $x_a$ and $x_b$, as follows:

$(1) \quad$ Find the points $-x_a$ and $x_b$ on the $x$-axis.
$(2) \quad$ Find the points $A$ and $B$ where the lines $x = -x_a$ and $x = x_b$ cut $P$.
$(3) \quad$ Lay a straightedge on the straight line joining $A$ and $B$ and locate its $y$-intercept $c$.

Then $x_a x_b$ can be read off from the $y$-axis as the position of $c$.


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 $m$ of the line segment joining $A B$ will be:

\(\ds m\) \(=\) \(\ds \frac {x_b^2 - x_a^2} {x_b - x_a}\) Equation of Straight Line in Plane: Point-Slope Form
\(\ds \) \(=\) \(\ds \frac {\paren {x_b - x_a} \paren {x_b + x_a} } {x_b - x_a}\) Difference of Two Squares
\(\ds \) \(=\) \(\ds x_b + x_a\) cancelling, $x_a \neq x_b$


From Equation of Straight Line in Plane: Slope-Intercept Form:

$y = \paren {x_b + x_a} x + c$

where $c$ denotes the $y$-intercept.


Substituting the coordinates of point $A = \tuple {x_a, x_a^2}$ for $\tuple {x, y}$:

\(\ds x_a^2\) \(=\) \(\ds \paren {x_b + x_a} x_a + c\)
\(\ds \leadsto \ \ \) \(\ds c\) \(=\) \(\ds x_a^2 - \paren {x_a + x_b} x_a\)
\(\ds \) \(=\) \(\ds x_a^2 - x_a^2 - x_b x_a\)
\(\ds \) \(=\) \(\ds -x_b x_a\)

$\blacksquare$


Sources