a hyperbola passing through (8,6) consists of all points whose distance from the origin is a constant more than its distance from the point (5,2). find the slope of the tangent line to the hyperbola at (8,6).

Answer 1

Let P(x,y) be any point on the parabola.
According to the stated condition
√( (x-8)^2 + (y-6)^2 ) - √( (x-5)^2 + (y-2)^2 ) = c, where c is that constant

we could find c by plugging in (8,6) , a point on the parabola, but there is no need for it

messy looking derivative:
(1/2)(( (x-8)^2 + (y-6)^2 )^(-1/2) (2(x-8) + 2(y-6)(dy/dx) + (1/2)(( (x-5)^2 + (y-2)^2 )^(-1/2) (2(x-5) + 2(y-2) dy/dx ) = 0

Very carefully, plug in x=8 and y=6, then find dy/dx
I will leave that onerous task for you.
Now you will have the slope and a point, piece of cake after that

Answer 2

All right! Thank you :)