Posted by **Shayne** on Wednesday, September 14, 2011 at 9:35pm.

Find an equation of the normal line to the parabola

y = x2 − 7x + 5

that is parallel to the line

x − 3y = 3.

- calculus -
**Reiny**, Wednesday, September 14, 2011 at 10:29pm
dy/dx = 2x-7 which is the slope of the tangent.

so the normal must have a slope of -1/(2x-7) or 1/(7-2x)

but that is supposed to be parallel to x-3y=3

that is, a slope of 1/3

1/(7-2x) = 1/3

3 = 7-2x

2x = 4

x = 2

when x=2, y = 4-14+5 = -5

so the normal has equation

x - 3y = c

with (2,-5) on it, so

2 + 15 = c

equation of normal is

x - 3y = 15

check: at (2,-5) slope should be -3

dy/dx = 2(2)-7 = -3

My answer for the equation of the normal is correct!!

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