00:01
Okay, what we want to show is that the equation of the tangent line to this hyperbola that is defined to be x squared over a squared minus y squared over b squared equal to one at x not comma y not is actually x not.
00:33
Over a squared times that x value minus y not over b squared times the y value equal to one okay so since we're talking about a tangent line you know we're going to have to be taking the derivative so let's take the derivative of with respect to x of that hyperbola equation and when we do um we get two x over a squared minus 2y over b squared times y prime equal to zero and so when we solve for y prime or the derivative of y with respect to x we get b squared x over a squared y and so now if we evaluate that derivative at the point x not y not we get b squared x not over a squared y not.
01:40
Okay, and so if we solve, if we write the equation for the line at that point, we get y minus y not is equal to that slope b squared x not over a squared y not times x minus x not.
01:59
Okay, and so now what we want to do is to go ahead and kind of simplify this out.
02:06
So we get y minus y -not is equal to b squared x -not times x over a squared times y -not minus b squared x -not squared over a squared y -not.
02:28
And then what we're going to do is don't like the denominator.
02:32
So we're going to multiply everything by a squared y -not.
02:35
And we get a squared y0, y minus a squared y not squared is equal to b squared x not x minus b squared x not squared...