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Given an ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$, $a > b>0$ find out the minimum area of its tangential parallelogram, and specify when this max area is achieved.

Pin Yin
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  • What is tangential parallel quadrilateral? Do you mean trapezium (two of the sides are parallel)? –  Mar 21 '17 at 05:11
  • Hey, sorry. I meant for parallelogram, I'll edit my question – Pin Yin Mar 21 '17 at 05:54
  • You might start by figuring out the equation for one side of the parallelogram. – amd Mar 21 '17 at 06:37
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    I think you want the minimum area. There is no maximum - consider a circle (special case of ellipse). We can draw parallelograms of arbitrary size around the circle (all of these will be rhombuses). The minimum area is $4ab$. Use the mapping $(x, y) \rightarrow \left(x, \frac{by}{a}\right)$ to translate the problem to the circle. –  Mar 21 '17 at 06:39
  • Thanks a lot @Muralidharan. Yes, it should be minimum. And you solution is perfect. And I found a way to visualize your solution, so it totally makes sense! – Pin Yin Mar 21 '17 at 17:23

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Thanks to @Muralidharan, the map $(x,y)\rightarrow (x, \frac{by}{a})$ would solve it. The map maps an ellipse to a circle, and the tangential parallelograms will remain as the tangential parallelograms under this map. Thus the minimum is achieved when it is a rec-tangle tangent to a circle gets linearly transformed. Thus the area is $4ab$.

Plus I found a perfect way of visualizing the above. Let's have a two plane intersected with certain dihedral angle, and then orthogonal projection of a circle with tangential parallelograms onto another plane is just a ellipse with a tangential parallelogram. And we could calculate the area by the theorem of orthogonal projections of plane area regions

Pin Yin
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