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Math Help - Max area for triangle in ellipse...

  1. #1
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    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..
    I uploaded the figure...
    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
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  2. #2
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    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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