00:01
We would define the tangent line to the curve given by the equation, y equals x over y plus a, at the point zero zero.
00:09
To do so, we're going to have to use implicit differentiation to obtain dydx or the derivative of a y with respect to x on this curve.
00:18
We can then use dydx, which is the change in y with a change in x or the slope, to find the slope at that point, and then plugging in our point and that d, y, d, d ,x, at the point into y equals mx plus b, we can solve it to b intercept, and find the equation for this tangent line.
00:33
So, before we get started, let's make this a little simpler.
00:37
If we multiply both sides by y plus a, this equation becomes y -square -plus -a -y equals x.
00:42
Now, applying them plus a differentiation to part one, differentiating with respect to x, our equation becomes d -d -x -y -squared plus a -y, equals d -d -x...