text: # Max area for triangle in ellipse...
  
  • Jun 27th 2010, 06:51 AM
  ice_syncer
  Max area for triangle in ellipse...
  This is the question :
  
  there is an ellipse of the form
  
  x^2/a^2 + y^2/b^2 = 1
  
  find max area of isosceles triangle inscribed in the ellipse with its vertex at one end of major axis..
  Now this is what I did.
  
  From the figure, I took the semi height as y and the length of base as x+a so the area will be A = y*(x+a)
  
  using the ellipse equation I got expression for y in terms of x and the constants a and b. then I did d/dx (A) = 0 to find the value of x, I got x = -a and x = 2a-1
  so that means the length is 2a + 1 and y = (b/a ) * (SQRT(a^2-x^2))
  and I get area as (b/a)((2a-1)^3/2) but the answer is b/a((3)^3/2)/4 ...
  
  I want to know whats wrong, I dont want to know any new method because I know one method mentioned in the book that is to take it in parametric form ie x=acost and y = bsint
  I want to know whats wrong with my method..
  Thanks
  • Jun 27th 2010, 07:30 AM
  skeeter
  $A = y(x+a)$
  
  $\displaystyle A = \frac{b}{a}\sqrt{a^2-x^2} \cdot (x+a)$
  
  $\displaystyle \frac{dA}{dx} = \frac{b}{a}\left(\sqrt{a^2-x^2} - \frac{x(x+a)}{\sqrt{a^2-x^2}}\right) = 0$
  
  $a^2-x^2 = x^2+ax$
  
  $0 = 2x^2 + ax - a^2$
  
  $0 = (2x-a)(x+a)$
  
  $x = \frac{a}{2}$ , $x = -a$
  
  throw out the solution $x = -a$ for the obvious result (triangle with $A = 0$ formed)
  
  $\displaystyle A = \frac{3\sqrt{3} \cdot ab}{4}$
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