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Question

Find the maximum area of an isosceles triangle inscribed in the ellipse x2a2+y2b2=1 with is vertex as at on end of the major axis.

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Solution

Area ΔAPP=12×pp×AM
=12×ab(2bsinθ)(aacosθ)
=ab[sinθ12sin2θ]
2A2θ=0
θ=2π3
2A2θ=ab[332]<0
A is maximum when θ=2π3=120o
Maximum valye of A=a(sin120o12sin240o)
=334ab.

1210664_1328107_ans_eee47990d1f5472eabef6f5c01406aa3.jpg

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