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A geometry student wants to draw a rectangle inscribed in the ellipse x² + 4y2 = 49.

What is the area of the largest rectangle that the student can draw?

User KarelHusa
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1 Answer

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Maximum area = 2ab

from ellipse eqn

x^2 + 4y^2 = 49

(x+4y)^2= 49

underoot both sides

x + 4y = 7 ---eqn1

from


\frac{ {x}^(2) }{ {a}^(2) } + \frac{ {y}^(2) }{{b}^(2) } = 1

(ab)^2 = (ax + by)^2

ab= ax + by

compare to eqn1

ab=7

then, area of the largest triangle will be 14 units.

i hope it's ryt

User Gatekeeper
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