# Minimum area ellipse passing through the vertices of an isosceles trapezoid

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.

• what have you tried yourself? Oct 16, 2021 at 8:41
• I have edited the question. Please check my update. Oct 16, 2021 at 9:51
• your answer seems correct and your method would appear to be the most direct way. Oct 16, 2021 at 15:37 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- $$a=AD=\sqrt{3^2+5^2}\approx 5.83$$.

• But AD = $\sqrt{3^2 + 5^2} = 5.830952$ while $a = 5.902508$ so I don't know what theorem are you talking about ? Oct 16, 2021 at 17:53
• @HosamHajjir, I said it could be a theorem not there is a theorem. May be there is such theorem. I will look and inform you if I find something. Oct 16, 2021 at 17:58
• I don't see any relation between $a$ and $AD$ or between $b$ and $OE$, I don't know where you came up with this. Oct 16, 2021 at 23:24
• @HosamHajjir, Accurate drawing shows the measures are closed to what you found. Oct 17, 2021 at 5:01