Question

Alpha and beta are eccentric angles of two points a and b on the ellipse x^2/a^2 + y^2/b^2=1.If p(acostheta,bsintheta) be any point on the same ellipse such that the area of triangle pab is maximum then prove that theta=alpha+beta/2

Answer 1

Answer:

Area =

∣

∣

∣

∣

∣

∣

∣

∣

acosθ

acosα

acosβ

bsinθ

bsinα

bsinβ

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

so, Area max

dθ

dA

=

∣

∣

∣

∣

∣

∣

∣

∣

asinθ

acosα

acosβ

bsinθ

bcosα

bsinβ

1

1

1

∣

∣

∣

∣

∣

∣

∣

∣

=0 (for maxime)

=−asinθ[bsinα−bsinβ]−bcosθ[acosα−acosβ]

=ab[−(cos(θ−α)+cos(θ−β))]=0

∴cos(θ−β)=cos(θ−α)

∴cos(−θ)=cos(θ)

⇒θ−β=−(θ−α) or θ−β=−(θ−α)

∴β=θ

θ=

2

α+β