Determine the minimum area of a parallelogram each of whose four sides are tangent to the ellipse x²/49 + y²/36 = 1.

Hi!

I try to find a analytical solution to the problem!

Let be ABCD the parallelogram.

Let be (x1,y1) tangent point from AB.

Let be (x1',y1') tangent point from CD.

I will prove that x1'=-x1 and y1'=-y1

The asociate equation for these 2 tangent points are :

xx1/49+yy1/36=1 and xx1'/49+yy1'/36=1

Because AB||CD  the 2 equation have same slope so x1/y1=x1'/y1' and because (x1,y1) and (x1',y1') are on the ellipse so x1'=-x1 and y1'=-y1.

So the tangent point are in fact

M(x1,y1) ; N(x2,y2) ; P(-x1,-y1) ; Q(-x2,-y2)

I can  prove that

the pairs of these points  (M,P); (N,Q); (A,C); (B,D) are collinear with the origin O.

So the area if [ABCD] = 4* [BOC].

I start to calculate the area of [BOC] and i find that

[BOC] = 49*36/(x1*y2-y1*x2)

and the area of [ABCD] = 4*49*36/(x1*y2-y1*x2)

After this point i have dificulties!!!

Because  the area of ABCD is minimium when the expresion (x1*y2-y1*x2) is maximum. These area is in fact the area of

2*[MOP] where the M and P are tangent point to the ellipse.

Is true that the maximum will be when is a rectangle or rombus but is only a intuition!!

So the intuition solution is :

The maximum value for [MOP] is 6*7 and the minimum value for [ABCD] is 4*49*36/6*7 = 4*7*6=(2*7)*(2*6)=14*12=168 like Brian solution.

Is just some calculus, not a sure solution!

 

Edited on January 6, 2008, 3:45 pm

Posted by Chesca Ciprian on 2008-01-06 15:43:23