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Question
Find the area of greatest rectangle that can be inscribed in an ellipse `x^2/"a"^2 + y^2/"b"^2` = 1
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Solution
Let ABCD be the rectangle of maximum area with sides AB = 2x and BC = 2y
Where C (x, y) is a point on the ellipse `x^2/"a"^2 + y^2/"b"^2` = 1 as shown in the Fig.6.3.
The area A of the rectangle is 4xy
i.e. A = 4xy which gives A2 = 16x2y2 = s ....(say)
Therefore, s = `16x^2 (1 - x^2/"a"^2)*"b"^2`
= `(16"b"^2)/"a"^2 ("a"^2x^2 - x^4)`
⇒ `"ds"/"dx" = (16"b"^2)/"a"^2 * [2"a"^2x - 4x^3]`
Again, `"ds"/"dx"` = 0
⇒ x = `"a"/sqrt(2)` nad y = `"b"/sqrt(2)`
Now, `("d"^2"s")/("dx"^2) = (16"b"^2)/"a"^2 [2"a"^2 - 12x^2]`
At x = `"a"/sqrt(2), ("d"^2"s")/("dx"^2)`
= `(16"b"^2)/"a"^2 [2"a"^2 - 6"a"^2]`
= `(16"b"^2)/"a"62 (-4"a"^2) < 0`
Thus at x = `"a"/sqrt(2)`, y = `"b"/sqrt(2)`
s is maximum and hence the area A is maximum.
Maximum area = 4.x.y
= `4 * "a"/sqrt(2) * "b"/sqrt(2)`
= 2ab sq units.
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