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 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
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
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 . Thus the point is . What is the equation of the normal there?
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