Asked by Al
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).
Answers
Answered by
Reiny
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
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
Answered by
Al
All right! Thank you :)