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 = 2a1
so that means the length is 2a + 1 and y = (b/a ) * (SQRT(a^2x^2))
and I get area as (b/a)((2a1)^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
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 = 2a1
so that means the length is 2a + 1 and y = (b/a ) * (SQRT(a^2x^2))
and I get area as (b/a)((2a1)^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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