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प्रश्न
Find the maximum area of an isosceles triangle inscribed in the ellipse `x^2/ a^2 + y^2/b^2 = 1` with its vertex at one end of the major axis.
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उत्तर
Ellipse, `x^2/a^2 + y^2/b^2 = 1`
Let P be a point (a cos θ, b sin θ) on the ellipse. APP' is an isosceles triangle.
PP' intersects the axis AA' of the ellipse at point M.
Area of ∆APP' A = `1/2 PP' xx AM`
`= 1/2 (2 b sin theta) xx (a - a cos theta)` ...[∵ PP' = 2 PM = 2b sin θ and AM = OA - OM = a - a cos θ]
`= 1/2 "ab" xx 2 sin theta (1 - cos theta)`
= ab sin θ (a - a cos θ)
= ab (sin θ - sin θ cos θ)
`= "ab"(sin θ - 1/2 * 2 sin θ cos θ)`
`= "ab" (sin theta - 1/2 sin 2 theta)`
On differentiating with respect to θ,
`(dA)/(d theta) = ab (cos theta - cos 2 theta)`
For highest to lowest, `(dA)/(d theta) = 0`
∴ ab(cos θ - cos 2θ) = 0
⇒ cos 2θ = cos θ
⇒ 2θ = 2π - θ
⇒ θ = `(2pi)/3`
Now, `(d^2A)/(d theta^2) = ab (- sin theta + 2 sin 2 theta)`
At, `theta = (2pi)/3 (d^2A)/(d theta^2) = ab (- sin (2pi)/3 + 2 sin (4pi)/3)`
`= ab [- (sqrt3/2) + 2 (- sqrt3/2)]`
`= ab (- sqrt3/2 - (2sqrt3)/2)`
`= (- 3 sqrt3)/2 ab < 0`
⇒ A is maximum when `theta = (2pi)/3 = 120^circ`
Maximum value of a = `ab (sin 120^circ - 1/2 sin 240^circ)`
`= ab [sqrt3/2 - 1/2 (- sqrt3/2)]`
`= ab(sqrt3/2 + sqrt3/4)`
`= (3sqrt3)/4`ab
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