posted by Jon on .
Find the equation of the tangent line and the normal line to the graph of the equation at the indicated point.
I need to show work, so answers formatted in this manner would be most appreciated. Thanks! :) :)
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
Thanks so much! :)
You're a lifesaver!