Question

2(x^2 + y^2)^2 = 25 ( x^2 - y^2 )

Answer 1

Answer:

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Answer 2

Given :   2(x² + y²)² = 25 ( x² - y² )

To find : Equation of tangent   at  point (-3 , 1)

Solution:

Complete Question find Equation of tangent at  point (-3 , 1)

2(x² + y²)² = 25 ( x² - y² )

d(2(x² + y²)² )/dx = d(25 ( x² - y² ))/dx

=> 4(x² + y²)(2x  + 2y.dy/dx)    = 25(2x  -  2y/dy/dx)

=>  4(x² + y²)(x  + y.dy/dx)  = 25(x  - y.dy/dx)

putting x = -3  & y = 1

=> 4(9 + 1)(-3  + dy/dx) = 25(-3 - dy/dx)

=> 40(-3 + dy/dx) = -75  - 25dy/dx

=> -120  + 40dy/dx  = -75 - 25dy/dx

=> 65dy/dx  = 45

=> dy/dx = 45/65

=> dy/dx =  9/13

y - 1  =  (9/13)(x  -  (-3))

=> 13y - 13  = 9x  + 27

=> 9x  - 13y  + 40  = 0

9x  - 13y  + 40  = 0 is the Equation of tangent at  point (-3 , 1) for 2(x² + y²)² = 25 ( x² - y² )

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