# How can I prove this property of an ellipse?

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.

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.

• We need more info. What are the center and radius of the circular arc above the ellipse? What are Q and V? Sep 8, 2021 at 3:10
• I've edited the picture. The circle, Q and V don't enter into the picture. Sep 8, 2021 at 4:33
• 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. Sep 8, 2021 at 6:05

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.