Isosceles triangle inscribed in an ellipse.Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse x 2 a 2 + y 2 b 2 = 1 with its (unique) vertex lying at one of the ends of the major axis of the ellipse.The three vertices of the triangle would be (a,0), (x,y), (x,-y) .The area of the triangle by Heron's formula is (1) A 2 = ( x − a ) 2 y 2 = ( x − a ) 2 b 2 ( 1 − x 2 a 2 ) . Hence d A d x = 0 ⟹ ( x − a ) 2 ( x + a 2 ) = 0..We have minimum at x = a and maximum at x = − a 2 .Substituting back in (1) and taking square roots on both the sides gives A = 3 a b 4 .The given answer is 3A.What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene?

Question

Isosceles triangle inscribed in an ellipse.Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse                     x        2                    a        2              +                    y        2                    b        2              =  1 with its (unique) vertex lying at one of the ends of the major axis of the ellipse.The three vertices of the triangle would be (a,0), (x,y), (x,-y) .The area of the triangle by Heron&#039;s formula is                     (1)                              A          2                =        (        x        −        a                  )          2                          y          2                =        (        x        −        a                  )          2                          b          2                          (          1          −                                                    x                2                                            a                2                                              )                .            Hence                     d        A                    d        x              =  0    ⟹    (  x  −  a      )    2        (    x    +                  a        2              )    =  0..We have minimum at   x  =  a and maximum at   x  =  −      a    2  .Substituting back in (1) and taking square roots on both the sides gives   A  =                              3                a        b            4      .The given answer is 3A.What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene?

Helena Howard · Accepted Answer

Step 1Everything is correct until you use the value of x to calculate the area. You should have       A    2    =  (  −            3      a        2        )    2        b    2    (  1  −      1    4    ) Step 2which will give you the correct answer   A  =            3              3              4    a  b

Awainaideannagi · Accepted Answer

Step 1We perform an orthogonal projection to map the ellipse to the unit circle.Let the maximal area of our isosceles triangle be       A   which we wish to find, and let   △  A  B  C be the isosceles triangle with maximal area inscribed in our unit circle. Since orthogonal projections preserve area ratios,   △  A  B  C is the projection of the triangle with area       A  . Note that since we wish to maximise the area,   △  A  B  C is simply an equilateral triangle with area             3              3              4  .Step 2Hence, by preservation of area ratios,                                           A                     Area of ellipse                                    =                              [            A            B            C            ]                     Area of circle                                            ⟹                          A                                    =                              3                          3                                            4            π                          ⋅        π        a        b                                          =                              3            a            b                          3                                4                     which is our answer.