Question Number 102980 by ajfour last updated on 12/Jul/20

Commented byajfour last updated on 12/Jul/20

Find  a/b  if regions 1, 2, 3, 4, 5  have equal areas.

Commented bymr W last updated on 12/Jul/20

this is not possible for ellipse  no matter which value a/b has.

Commented bymr W last updated on 12/Jul/20

if this is true, say for a certain  value of a/b, then it must be also true  when we shrink the ellipse in  x−direction in ratio b/a to a circle  with radius r=b.  that means each of the three shaded  segments of the circle must have   fifth of the area of the circle. but  this is not possible, since  area of segement =(b^2 /2)(((2π)/3)−((√3)/2))≠((πb^2 )/5)

Commented bymr W last updated on 12/Jul/20

Commented byajfour last updated on 12/Jul/20

′′  then it must be also true  when we shrink the ellipse ′′       may not necessarily be true       Sir....

Commented bymr W last updated on 12/Jul/20

when an ellipse is shrinked in  x−direction in ratio (b/a), the area of  each part is shrinked in the same  ratio (b/a):  A_1  →A_1 ^′ =(b/a)A_1   A_2  →A_2 ^′ =(b/a)A_2   ...  therefore if A_1 =A_2 =A_3 =..., we get  also A_1 ′=A_2 ′=A_3 ′=...

Answered by ajfour last updated on 12/Jul/20

A_1 =((2b)/a)∫_h ^(  a) (√(a^2 −x^2 )) dx        = ((2b)/a){(h/2)(√(a^2 −h^2 ))+(a^2 /2)sin^(−1) (h/a)}∣_h ^a      =((2b)/a){((πa^2 )/4)−(h/2)(√(a^2 −h^2 ))−(a^2 /2)sin^(−1) (h/a)}  A_2 =(k/2)(h+a)  ,   k=b(√(1−(h^2 /a^2 )))  A_1 = A_2  =((πab)/5)        gives h and  (a/b)        ((2b)/a){((πa^2 )/4)−(h/2)(√(a^2 −h^2 ))−(a^2 /2)sin^(−1) (h/a)}            =  ((b(h+a))/2)(√(1−(h^2 /a^2 )))  let  (h/a)= p  ⇒  (π/2)−p(√(1−p^2 ))−sin^(−1) p              = (((p+1))/2)(√(1−p^2 ))   = (π/5)     Yes  Sir,  follow you only now!

Commented bymr W last updated on 12/Jul/20

you can make A_1 =A_2  or A_1 =A_3  or  A_2 =A_3  for any ellipse, but never  A_1 =A_2 =A_3 .

Commented byajfour last updated on 12/Jul/20

A_1 =A_2    gives h for any a and b;  while    A_1 =A_2 =((πab)/5)   for  specific     (a/b) .

Commented bymr W last updated on 12/Jul/20

yes, you can get A_1 =A_2  for any  ellipse, but never A_1 =A_2 =((πab)/5),  because A_1 =A_2 =((πab)/5) means the same  as A_1 =A_2 =A_3 =((πab)/5), which is not  possible.