Find the equation of the tangent line and the normal line to the graph of the equation at the indicated point.

Question

x^2-y^2=16, (5,3)

I need to show work, so answers formatted in this manner would be most appreciated. Thanks! :) :)

Answer 1

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

Answer 2

Thanks so much! :)

You're a lifesaver!