Question

~Find the maximum area of an isosceles triangle inscribed in the ellipse
x^2/16+y^2/9=1
  with its vertex at one end of the major axis.

Answer 1

Given :  an isosceles triangle inscribed in the ellipse x²/16 + y²/9 = 1

To Find :   maximum area of an isosceles triangle

Solution:  

x²/16 + y²/9 = 1

lets take one end of the major axis as A  = ( 4 , 0)

Points of B & C as ( x , -y)  and (x , y)

Base = 2y

Height = ( 4 - x)

Area = (1/2) 2y ( 4 - x)

A = (4 - x) y

x²/16 + y²/9 = 1 =>  9x² + 16y² = 144  => 16y² = 144 - 9x²  =>  y = (1/16)√144 - 9x²

= (3/4)√4² - x²

A = (4 - x)  (3/4)(√4² - x² )

dA/dx = (3/4) (  (4 - x) (-2x/ 2√4² - x²)  + (-1)(√4² - x² ))

= (3/4) (1/√4² - x²)( -4x + x²  - (4² - x²) )

= (3/4) (1/√4² - x²)( 2x² - 4x - 16 )

=  (3/2) (1/√4² - x²)(  x² - 2x - 8 )

x² -2x - 8 = 0

=>(x - 4)(x + 2) = 0

=> x = 4  , x = - 2

x = -4 not possible

Hence x = 2

A = (4 - x)  (3/4)(√4² - x² )

= 6(3/4)√12

= 9√3

maximum area of an isosceles triangle = 9√3 sq units

Learn More:

A triangle ABC is inscribed in the ellipse 2x2 + y2 = 18. If B is (1,4 ...

https://brainly.in/question/28065670

https://brainly.in/question/28113721