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.

Lorelei Patterson

Lorelei Patterson

Answered question

2022-07-18

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?

Answer & Explanation

Helena Howard

Helena Howard

Beginner2022-07-19Added 12 answers

Step 1
Everything 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 2
which will give you the correct answer A = 3 3 4 a b
Awainaideannagi

Awainaideannagi

Beginner2022-07-20Added 5 answers

Step 1
We 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 2
Hence, 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.

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