martes, 3 de enero de 2017

1102. Ellipses between hyperbolae

    I had drawn on the chalkboard the two conjugated equilateral hyperbolae  xy = ± 1  and it was easy to calculate the area of the circle that is tangent to the four hyperbola branches... Then Pepe Chapuzas interrupted me...

    Dear Teacher:
    My intuition tells me that all the ellipses that are tangent to the four branches have the same area. Am I wrong?
    Bravo for the intuition of Pepe! Was he right? Loli Palique answered...

    Dear Teacher:
    Intuition guides me often, but I'm aware that intuition often deceives... The first thing I'm going to do is calculate the area of the circle... Since its center is (0, 0) and the vertices of the hyperbolae are (±1, ±1), so its radius is

(12+12)  =  2
and its area is
πr2  =  2π 

    An ellipse with center (0,0) and axes  x = 0  and  y = 0  has this equation:

x2/a2 + y2/b2 = 1 

    Whether the area of that ellipse  πab  were  2π  too, then  b = 2/a . In such case, the intersection  of this ellipse and the hyperbola branch  y = 1/x  on the first quadrant  (x>0, y>0)  shall furfill

x2/a2 + a2/4/x2  =  1
x2/a2 + a2/4/x2    1  =  0
(x/a  a/2/x)2  =  0
x/a  =  a/2/x
x2  =  a2/2
x  =  a/2
y  =  2/a

    So, there is only one instersection point, that is, the ellipse is tangent to that branche and, in the same way, to the other branches... 
    Of course, the ellipse and the hyperbola have the same slope in their tangency points...
    Calculating derivatives with the ellipse equation...

(x2/a2 + ay2/4) '  =  2x/a2 + ay y'/2  =  0
y'  =  4x/a4/y  =  2/a2

    Calculating derivatives with the hyperbola equation...

(x y) '  =   y + x y'  =  0
y'  =  −y/x  =  2/a2

    Loli ended by saying...

    Dear Teacher:
    My intuition tells me that this proposition is true for conjugated hyperbolae even if they aren't equilateral.

    Is she right?

No hay comentarios:

Publicar un comentario