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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.

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    $\begingroup$ what have you tried yourself? $\endgroup$ Oct 16, 2021 at 8:41
  • $\begingroup$ I have edited the question. Please check my update. $\endgroup$ Oct 16, 2021 at 9:51
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    $\begingroup$ your answer seems correct and your method would appear to be the most direct way. $\endgroup$ Oct 16, 2021 at 15:37

1 Answer 1

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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- $a=AD=\sqrt{3^2+5^2}\approx 5.83$.

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  • $\begingroup$ But AD = $\sqrt{3^2 + 5^2} = 5.830952 $ while $a = 5.902508$ so I don't know what theorem are you talking about ? $\endgroup$ Oct 16, 2021 at 17:53
  • $\begingroup$ @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. $\endgroup$
    – sirous
    Oct 16, 2021 at 17:58
  • $\begingroup$ 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. $\endgroup$ Oct 16, 2021 at 23:24
  • $\begingroup$ @HosamHajjir, Accurate drawing shows the measures are closed to what you found. $\endgroup$
    – sirous
    Oct 17, 2021 at 5:01

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