Multiplication using Parabola

Theorem

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, $x_a \ne x_b$

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

• 1948: W.W. Sawyer: Mathematics In Theory and Practice: On Being your Own Teacher: Labour Saving Devices