# equation of normal to curve

• Apr 6th 2012, 07:50 AM
jessm001
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
• Apr 6th 2012, 07:59 AM
Plato
Re: equation of normal to curve
Quote:

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 $\displaystyle \tfrac{-1}{3}$ then the slope of the tangent at the same point is $\displaystyle 3$.
At what point(s) is the slope of the tangent $\displaystyle 3~?$
• Apr 6th 2012, 08:02 AM
jessm001
Re: equation of normal to curve
It doesn't say, that's why I got confused, I thought I was missing something
• Apr 6th 2012, 08:08 AM
princeps
Re: equation of normal to curve
Quote:

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 .
• Apr 6th 2012, 08:10 AM
jessm001
Re: equation of normal to curve
Thanks a lot :)
• Apr 6th 2012, 08:27 AM
Plato
Re: equation of normal to curve
Quote:

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 $\displaystyle y'=2x=3$. Thus the point is $\displaystyle \left( {\tfrac{3}{2},\tfrac{{ - 7}}{4}} \right)$.
What is the equation of the normal there?