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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