# Thread: Equations of tangents through point (curve)

1. ## 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...

2. 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...

$\displaystyle f'(x)=\frac{-1}{2(8-x)^{1/2}} \quad \leftarrow \quad \text{Right !}$

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

Find $\displaystyle f'(5)$

Then, the equation of the tangent is :

$\displaystyle y=f'(5)(x-5)+\underbrace{f(5)}_{=2}$

Is it more clear ?