Thread: Equations of tangents through point (curve)

Determine the exact equations of the tangents to the curve which pass through point P

curve = (8-x)^(1/2)
point = 5,2

I'd like to see how you would solve this. What I tried to do is set the variables of the point as: a, (8-a^(1/2))
and the slope would be
-1 /( 2(8-x)^(1/2) )

Is this right? Putting this into the y-y0 = m (x-x0) seems to get messy...

Hello,

Originally Posted by theowne

Determine the exact equations of the tangents to the curve which pass through point P

curve = (8-x)^(1/2)
point = 5,2

I'd like to see how you would solve this. What I tried to do is set the variables of the point as: a, (8-a^(1/2))
and the slope would be
-1 /( 2(8-x)^(1/2) )

Is this right? Putting this into the y-y0 = m (x-x0) seems to get messy...

Your post is already messy



You want it at the point (5,2), that is to say x=5.

Find


Then, the equation of the tangent is :



Is it more clear ?