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Find the area of greatest rectangle that can be inscribed in an ellipse abx2a2+y2b2 = 1

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प्रश्न

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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उत्तर


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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अध्याय 6: Application Of Derivatives - Solved Examples [पृष्ठ १२९]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 12
अध्याय 6 Application Of Derivatives
Solved Examples | Q 16 | पृष्ठ १२९

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