Posted by **Annie** on Friday, October 23, 2009 at 10:36pm.

Find the equation of the normal to the curve y = 3x^2 - 5x that is perpendicular to the line y = x + 4.

- Maths -
**bobpursley**, Friday, October 23, 2009 at 10:46pm
Find the slope of the tangent first.

the perpendicular to the line x+4 has a slope of -1.

so the normal to the curve has the same slope.

y'=6x-5 and then the perpendicular to this must have a slope of -1/6x

so -1/6x=-1 or x= 1/6

y=mx+b slope is -1

y=-x+b but the point of tangencey is x=1/6

y=-1/6+b

Now for y, in the curve y=3x^2-5x, when x is 1/6, then y= 3/36-30/36=-27/36

so solve for b.

All this depends on my intrepretation of normal to the curve that is perpendicular to the line...

That is most unusual wording.

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