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^2-y^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)/(x-5) = slope = 5/3

y-3 = (5/3)(x-5) = 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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