Max area for triangle in ellipse...

**ice_syncer** · June 27th 2010, 06:51 AM

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

**skeeter** · June 27th 2010, 07:30 AM

$A = y(x+a)$

$\displaystyle A = \frac{b}{a}\sqrt{a^2-x^2} \cdot (x+a)$

$\displaystyle \frac{dA}{dx} = \frac{b}{a}\left(\sqrt{a^2-x^2} - \frac{x(x+a)}{\sqrt{a^2-x^2}}\right) = 0$

$a^2-x^2 = x^2+ax$

$0 = 2x^2 + ax - a^2$

$0 = (2x-a)(x+a)$

$x = \frac{a}{2}$ , $x = -a$

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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