# Points of Normal line

1. Oct 27, 2009

### synergix

1. The problem statement, all variables and given/known data

Find the points where the normal line to the curve
y = sqrt(x - 1)
is parallel to the line y = 1 - 2x.

3. The attempt at a solution

m=-2

y' = 1/2sqrt(x-1) = m = -2

-4sqrt(x-1)=1

sqrt(x-1) = - 1/4

x= 1/16 + 1

x= 17/16
so find y
y = sqrt((17/16)-1))

(17/16, 1/4)
this is what I think I have to do but I know I am wrong just not sure why my instructor did not explain this question very well

2. Oct 27, 2009

### Dick

You've made a credible attempt, if not correct, if the problem had been find the tangent line to the curve parallel to the line 1-2x. But it doesn't say that. It's asking you where the slope of the NORMAL line is -2. What's the relation between the slope of the tangent line and the normal line?

3. Oct 27, 2009

### synergix

the relationship between the the slope of the tangent line and the slope of the normal line is that the tangent will be the inverse negative of the slope of the normal line.

so slope of the normal line= -1(-2-1)

1/2sqrt(x-1) = m = 1/2

sqrt(x-1) = 1

x = 2

4. Oct 27, 2009

### synergix

Thanks! shouldn't be any trouble now.

5. Oct 27, 2009

### Dick

Wow, you've cleverly turned this completely upside down. Are you trying to confuse me? No, the slope of the normal line is supposed to be -2. It's supposed to be parallel to 1-2x. That's -2. What's the slope of the normal line to y=sqrt(x-1)?? Equate them.

6. Oct 27, 2009

### Dick

What you did is find the normal slope to the line and equate it to the tangent direction of the curve. That's sort of backwards, but if you understand why that works, it's ok with me.

7. Oct 27, 2009

### synergix

Haha, damn now I am confused. The slope of the normal line to y=sqrt(x-1) is -2 at the points that I must find. So I need to know what does x equal when the derivative of [sqrt(x-1)]' = -2 that makes sense to me but it must be wrong because that's what I already did.

8. Oct 27, 2009

### synergix

Ok would it work? I just learned this stuff yesterday and my teacher didn't really explain it very well. So maybe you have a more straight forward way.

9. Oct 27, 2009

### Dick

The slope of the tangent line is 1/(2sqrt(x-1)). So the slope of the normal line is -2sqrt(x-1). Same reasoning you used for the line. You want that to be -2. So you want -2=(-2)sqrt(x-1). It's the same equation you just solved. What you did is perfectly ok. Just want to make sure you understand why.

10. Oct 27, 2009

### synergix

OH, OK I get it now. thank you.