Question

7. Prove that the acute angle between two conjugate diameters of an ellipse is minimum

when they are equal.​

Answer 1

Prove that the acute angle between two conjugate diameters of an ellipse is minimum when they are equal.​

Step-by-step explanation:

Let BQ and BE be the conjugate semi-diameters,

then

(  BQ⁻ -   BE⁻ )² = BQ² + BE² - 2BQ⁻.BE⁻

2BQ.BE =  a²+b²  -  (  BQ⁻ -   BE⁻ )²

The area of parallelogram

=ab = BQ⁻.BE⁻ sin(∠PCD)

2ab = [ a²+b² - (  BQ⁻ -   BE⁻ )² ]sin(∠QBE)

sin ( ∠QBE) = 2ab/[ a²+b² - (  BQ⁻ -   BE⁻ )² ]

For ∠QBE to be least  [ a²+b² - (  BQ⁻ -   BE⁻ )² ] must be greatest and this is only possible when BQ⁻= BE⁻

The diamters are equi-conjugate

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