# Find the slope of the tangent line to the curve (a lemniscate) `2(x^2+y^2)^2 = 25(x^2-y^2)` at the point (3,-1). m=?

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### 2 Answers

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

`2(x^4 + 2x^2 y^2 + y^4)= 25x^2 -25y^2 = 0`

`2x^4 + 4x^2 y^2 + 2y^4 = 25x^2 -25y^2 = 0`

`8x^3 +4(2xy^2 +2x^2 yy') + 8yy' = 50x -50y `

`8x^3 +8xy^2 +8x^2 yy' + 8yy' = 50x -50yy' `

`==> 8x^2yy' +8yy' +50yy' = 50x -8x^3 -8xy^2`

`==> y'(8x^2y +8y +50y) = (50x -8x^3 -8xy^2)`

`==> y' = (50x-8x^3-8xy^2)/(8x^2 y +58y)`

`==> y'(3,-1) = (50*3 -8(3^3) - 8(3)(-1))/ (8(3^2)(-1) + 58(-1))`

`==> y'(3,-1) = (150 -216 +24)/(-72 -58)`

`==> y'(3,-1) = -42/-130 = 21/65`

`==> m= 21/65`

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Expert but you made a mistake at the first line, you forgot the squared