# Thread: equation of normal to curve

1. ## equation of normal to curve

Find the equation of the normal to y = x^2 - 4 which is parallel to x + 3y - 1 = 0

I found that the gradient of the normal is -1/3 if this is correct, i don't know where to go from here .. if someone could show me the steps please

2. ## Re: equation of normal to curve

Originally Posted by jessm001
Find the equation of the normal to y = x^2 - 4 which is parallel to x + 3y - 1 = 0
I found that the gradient of the normal is -1/3 if this is correct
If the slope of the normal is $\tfrac{-1}{3}$ then the slope of the tangent at the same point is $3$.
At what point(s) is the slope of the tangent $3~?$

3. ## Re: equation of normal to curve

It doesn't say, that's why I got confused, I thought I was missing something

4. ## Re: equation of normal to curve

Originally Posted by jessm001
Find the equation of the normal to y = x^2 - 4 which is parallel to x + 3y - 1 = 0

I found that the gradient of the normal is -1/3 if this is correct, i don't know where to go from here .. if someone could show me the steps please
I think that there is a lack of informations...coordinates of the point that belongs to the curve should be given .

Thanks a lot

6. ## Re: equation of normal to curve

Originally Posted by princeps
I think that there is a lack of informations...coordinates of the point that belongs to the curve should be given .
That is not so. There is no lack of information.
Slope is $y'=2x=3$. Thus the point is $\left( {\tfrac{3}{2},\tfrac{{ - 7}}{4}} \right)$.
What is the equation of the normal there?