Is an Ellipse a Circle?

Putting the equation of the ellipse into standard form

$\dfrac{x^2}{ (5/3)^2} + \dfrac{y^2}{ (5/4)^2 }= 1$

Let $x' = \dfrac{3}{5} x$ , and $y' = \dfrac{4}{5} y$ ,

then

$x'^2 + y'^2 = 1$

$x = 1, y = 1$ , corresponds to $x' = 3/5, y' = 4/5$

The other two vertices are vertices of the equilateral triangle inscribed in the unit circle with one vertex at (3/5, 4/5)

so, $(x'_2, y'_2) = R(+120^{\circ}) (3/5, 4/5) = ( 3/5 \cos(120^{\circ}) - 4/5 \sin(120^{\circ}), 3/5 \sin(120^{\circ}) + 4/5 \cos(120^{\circ}))$

and $(x'_3, y'_3) = R(-120^{\circ}) (3/5, 4/5) = ( 3/5 \cos(120^{\circ}) + 4/5 \sin(120^{\circ}), -3/5 \sin(120^{\circ}) + 4/5 \cos(120^{\circ}))$

hence,

$y_1 y_2 y_3 = y_2 y_3 = (5/4)^2 y'_2 y'_3 = (5/4)^2 ( (4/5)^2 (1/4) - (3/5)^2 (3/4) ) = -11/64 =\boxed{ -0.171875}$