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

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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

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If the slope of the normal is then the slope of the tangent at the same point is .
At what point(s) is the slope of the tangent

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

Quote:

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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

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I think that there is a lack of informations...coordinates of the point that belongs to the curve should be given .

Thanks a lot :)

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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 .

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That is not so. There is no lack of information.
Slope is . Thus the point is .
What is the equation of the normal there?