I am reading Maxwell's Matter and Motion and he has this construction as a step in deriving Newton's law of Gravitation from Kepler's First Law.

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In this construction $SU$ is equal to the ellipse's major axis $AB$, and $PZ$ is the perpendicular bisector to $HU$. Maxwell states that $HZ \times SY=b^2$, with $b$ being the length of the ellipse's semiminor axis. I can see how this is valid when $HZ=SY$ and I have an idea of how to derive it analytically, but I would like to know how to derive it from properties of the circle and the ellipse, etc. using classical geometry.

  • $\begingroup$ We need more info. What are the center and radius of the circular arc above the ellipse? What are Q and V? $\endgroup$ Sep 8, 2021 at 3:10
  • $\begingroup$ I've edited the picture. The circle, Q and V don't enter into the picture. $\endgroup$ Sep 8, 2021 at 4:33
  • 1
    $\begingroup$ I think one way to approach this diagram is via the tangent line directly. Given a point $P$ on the ellipse and its tangent line, we can reflect the points $H$ across said line to get the point $U$, with $Z$ being the projection of either point onto the tangent line. (Similarly $Y$ is the projection of $S$ onto the tangent line.) One then argues that $\overline{UP}=\overline{HP}$ and therefore $\overline{SU}=\overline{SP}+\overline{PU}=\overline{SP}+\overline{HP}=\overline{AB}$ by the definition of an ellipse. $\endgroup$ Sep 8, 2021 at 6:05

1 Answer 1


Extend $HP$ to meet $SY$ at point $Q$.

$HUQS$ is an isosceles trapezoid and hence a cyclic quadrialteral. Applying Ptolemy's Theorem gives $HZ\cdot SY=\frac {1}{4}\left(AB^{2}-SH^{2}\right)$, which is the length of the semiminor axis.


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