Expert: Sherman D. - 11/19/2008

Question
Hello,
I have a question about implicit differentiation: How can I 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)??
Thank you for your help!

Answer
2(x^2 + y^2）^2 = 25(x^2 − y^2)

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

4(x^2 + y^2)(2x + 2y(dy/dx)) = 25(2x - 2y(dy/dx))

8x(x^2 + y^2) + 8y(x^2 + y^2)(dy/dx) = 50x - 50y(dy/dx)

8y(x^2 + y^2)(dy/dx) + 50y(dy/dx) = -8x(x^2 + y^2) + 50x

if you divide everything by 2, you get

4y(x^2 + y^2)(dy/dx) + 25y(dy/dx) = -4x(x^2 + y^2) + 25x

(dy/dx)(4y(x^2 + y^2) + 25y) = -4x(x^2 + y^2) + 25x

(dy/dx) = (-4x(x^2 + y^2) + 25x)/(4y(x^2 + y^2) + 25y))

now plug (-3,-1) in for x and y

(dy/dx) = (-4(-3)((-3)^2 + (-1)^2) + 25(-3)) / (4(-1)((-3)^2 + (-1)^2) + 25(-1))

(dy/dx) = (12(9 + 1) - 75)/(-4(9 + 1) - 25)

(dy/dx) = (120 - 75)/(-40 - 25)

(dy/dx) = (45/-65)

(dy/dx) = -9/13

Thats the slope of the tangent, and to find the equation

-1 = (-9/13)(-3) + b
-1 = (27/13) + b
-1 - (27/13) = b
b = (-13 - 27)/13
b = (-40/13)

y = (-9/13)x - (40/13)

so

Slope = (-9/13)
Equation = y = (-9/13)x - (40/13)

here is a website that has the exact same equation, only the coordinates are positive instead of negative. However it doesn't show you the work step by step.

www.libraryofmath.com/implicit-differentiation.html