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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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Answer» Let any point on the ellipse `x^(2)/a^(a)+y^(2)/b^(2) = 1" be "P (a cos theta, b sin theta)`. ` (##NTN_MATH_XII_C06_E04_270_S01.png" width="80%"> Draw `PM bot OX` from P and produce it to meet the ellipse at Q, then `Delta APQ` is an isosceles triangle, let its area be S, then `S = 2 xx 1/2 xx AM xx MP` ` = (OA - OM) xx MP` ` = (a - a cos theta)* b sin theta` ` rArr S = ab (sin theta-1/2 sin 2 theta)` ` = ab ( sin theta - 1/2 sin 2 theta)` ` rArr (dS)/(d theta) = ab (cos theta- cos 2 theta)` Again differentiate w.r.t.` theta`, `(d^(2)S)/(d theta^(2)) = ab ( - sin theta + 2 sin 2 theta)` For maxima/minima ` (dS)/(d theta) = 0 ` ` rArr cos theta = cos 2 theta rArr 2 theta = 2 pi - 0 rArr theta =(2pi)/3 at theta=(2pi)/3`, `((d^(2)S)/(d theta^(2)))_(theta=(2pi)/3)= ab [-sin(2pi)/2 + 2 sin (2 xx(2pi)/3)]` ` = ab (-sqrt3/2 - (2sqrt3)/2)=ab ((-3sqrt3)/2)` ` = (-3sqrt3 ab)/2 lt 0 ` ` :." S is maximum at "theta = (2pi)/3`, and maximum value of S `= ab (sin (2pi)/3 - 1/2*2 sin.(2pi)/3cos. (2pi)/3)` ` = ab (sqrt3/2 + sqrt3/2 xx 1/2)` `= ab ((2sqrt3+sqrt3)/4)= (3sqrt3)/4` ab sq. units . |
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