Posted by Jon on Sunday, November 11, 2007 at 12:21pm.
Find the equation of the tangent line and the normal line to the graph of the equation at the indicated point.
x^2y^2=16, (5,3)
I need to show work, so answers formatted in this manner would be most appreciated. Thanks! :) :)

Calculus  drwls, Sunday, November 11, 2007 at 12:30pm
For the tangent line slope, you want dy/dx.
In your case you can differentiate both sides of the equation with respect to x, treating y as a function of x.
2x  2 dy*dy/dx = 0
dy/dx = x/y = 5/3
For the equation of the line, write it in the form
(y 3)/(x5) = slope = 5/3
y3 = (5/3)(x5) = 5/3 x 25/3
y = (5/3)x 16/3

Calculus  Jon, Sunday, November 11, 2007 at 12:50pm
Thanks so much! :)
You're a lifesaver!
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