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if x^2+y^2 = t - 1/t and x^4 + y^4 = t^2...

if `x^2+y^2 = t - 1/t` and` x^4 + y^4 = t^2 + 1/t^2 `then prove that `dy/dx = 1/(x^3y)`

A

`(1)/(x^(2)y^(2))`

B

`(1)/(xy^(3))`

C

`(1)/(x^(2)y^(2))`

D

`(1)/(x^(3)y)`

Text Solution

Verified by Experts

The correct Answer is:
D
Given ` x ^ 2 + y^ 2 = t - ( 1 ) /( t ) ` and ` x ^ 4 + y ^ 4 = t ^ 2 + ( 1 ) /( t ^ 2 ) `
`rArr x ^ 4 + y^ 4 + 2x ^ 2 y ^ 2 = t ^ 2 + ( 1 )/( t ^ 2 ) - 2 `
` rArr x ^ 4 + y ^ 4 + 2 x ^ 2 y ^ 3 = x ^ 4 + y ^ 4 - 2 `
` rArr x ^2y ^ 2+ 1 = 0 `
` y^2 = ( - 1 ) /( x ^ 2 ) `
On differentiating w.r.t x we get
` 2y ( dy ) /(dx ) = ( 2 )/( x ^ 3 ) `
` rArr (dy ) /( dx) = ( 1 )/( x ^ 3 y ) `
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