An isosceles trapezoid has its four vertices as follows: $A(0, 0), B(10, 0), C(7, 5), D(3, 5)$. I want to find the ellipse passing through the four vertices and having the minimum possible area. What is the equation of this ellipse ?
What I have tried:
From symmetry, and orientation of the trapezoid, the center of the ellipse is at $(5, y_0)$ and its equation is
$ \dfrac{(x - 5)^2}{a^2} + \dfrac{(y - y_0)^2 }{b^2 } = 1 $
Since $(0, 0)$ is on the ellipse, then
$ \dfrac{25}{a^2} + \dfrac{ y_0^2}{b^2 } = 1 $
Since $(3, 5)$ is on the ellipse, then
$\dfrac{4}{a^2} + \dfrac{ (5-y_0)^2}{b^2} = 1 $
which are two equation in three unknowns, and they are linear in $\dfrac{1}{a^2} $ and $\dfrac{1}{b^2} $, hence, it can be solved readily to obtain:
$\dfrac{1}{a^2} = \dfrac{ (5- y_0)^2 - y_0^2 }{ 25(5 - y_0)^2 - 4 y_0^2 }$
$\dfrac{1}{b^2} = \dfrac{21}{ 25(5 - y_0)^2 - 4 y_0^2 } $
Therefore, the area of the ellipse is
$A = \pi a b = \pi \dfrac{ 25(5- y_0)^2 - 4 y_0^2 }{\sqrt{21} \sqrt{25 - 10 y_0} }$
And now to find the minimum area, I differentiate $A$ with respect to $y_0$
$\dfrac{d A}{d y_0} = 0 $ implies that
$ (-50(5- y_0) - 8 y_0)( \sqrt{25 - 10 y_0} ) + \dfrac{5}{\sqrt{25 - 10 y_0} } (625 - 250 y_0 + 21 y_0^2) = 0 $
Multiplying through by $\sqrt{25 -10 y_0} $, I get,
$ (-250 + 42 y_0) (25 - 10 y_0) + 5 ( 625 - 250 y_0 + 21 y_0^2 ) = 0$
Which simplifies to the following quadratic equation,
$ 315 y0^2 -2300 y_0 + 3125= 0$
The solutions of which are: $1.804809$ and $5.496778$
The second one is extraneous. Therefore,
$a^2 = \dfrac{ 25(5 - y_0)^2 - 4 y_0^2 }{ (5- y_0)^2 - y_0^2 }$
$b^2 = \dfrac{ 25(5 - y_0)^2 - 4 y_0^2 }{21} $
And these give: $a = 5.902508, b = 3.396089 $
So that the equation of the ellipse is now fully specified.
I wonder whether there is a shorter and more direct way to solving this problem.
Comment: Your calculation shows that if:
$a=AD=BC=5.83$
and:
$b=r=OF=OE$
then the area of circum - elipse is minimum.It could be a theorem. So easier way is:
1- finding the measure of OF=OE=b through a system of two equations in term of $y_o$, made by distances from O and lines AD and BC, which must give $y_o\approx 2$ then $b=OE=OF\approx 3.4$.
2- $a=AD=\sqrt{3^2+5^2}\approx 5.83$.