Find the maximum area of an isosceles triangle inscribed in the ellipse x 2 a 2 + y 2 b 2 = 1.Find the maximum area of an isosceles triangle inscribed in the ellipse x 2 / a 2 + y 2 / b 2 = 1. My teacher solved it by considering two arbitrary points on the ellipse to be vertices of the triangle, being ( a cos ⁡ θ , b sin ⁡ θ ) and ( a cos ⁡ θ , − b sin ⁡ θ ). (Let's just say θ is theta) and then proceeded with the derivative tests(which i understood) But, he didn't indicate what our θ was,and declared that these points always lie on an ellipse. Why so? And even if they do, what's the guarantee that points of such a form will be our required vertices? One more thing, I'd appreciate it if you could suggest another way of solving this problem.

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Find the maximum area of an isosceles triangle inscribed in the ellipse             x      2              a      2        +            y      2              b      2        =  1.Find the maximum area of an isosceles triangle inscribed in the ellipse       x    2        /        a    2    +      y    2        /        b    2    =  1. My teacher solved it by considering two arbitrary points on the ellipse to be vertices of the triangle, being   (  a  cos  ⁡  θ  ,  b  sin  ⁡  θ  ) and   (  a  cos  ⁡  θ  ,  −  b  sin  ⁡  θ  ). (Let&#039;s just say   θ is theta) and then proceeded with the derivative tests(which i understood) But, he didn&#039;t indicate what our   θ was,and declared that these points always lie on an ellipse. Why so? And even if they do, what&#039;s the guarantee that points of such a form will be our required vertices? One more thing, I&#039;d appreciate it if you could suggest another way of solving this problem.

emerhelienapj · Accepted Answer

Step 1If we believe the symmetry argument, it is a simple matter to optimize area:      A    1    =  a  b  sin  ⁡  θ  (  1  +  cos  ⁡  θ  )      A    2    =  a  b  (  1  +  sin  ⁡  θ  )  cos  ⁡  θ  0  =            ∂              A        1                    ∂      θ        =  a  b  [  cos  ⁡  θ  (  1  +  cos  ⁡  θ  )  −      sin    2    ⁡  θ  ]  =  a  b  [      cos    2    ⁡  θ  −      sin    2    ⁡  θ  +  cos  ⁡  θ  ]  0  =            ∂              A        2                    ∂      θ        =  a  b  [      cos    2    ⁡  θ  −  sin  ⁡  θ  (  1  +  sin  ⁡  θ  )  ]  =  a  b  [      cos    2    ⁡  θ  −      sin    2    ⁡  θ  −  sin  ⁡  θ  ]Step 2Solutions to the first equation include a minimum at   θ  =  π and maxima at   θ  =  ∓      π    3  .Interesting, we get solutions aligned with the x-axis and the y-axis!You can certainly solve the problem taking three arbitrary points on the ellipse, constraining them to lie on iscoceles triangles, and then use Lagrange multipliers. The solutions will turn out to be those found above.

Find the maximum area of an isosceles triangle inscribed in the ellipse x^2/a^2+y^2/b^2=1.

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