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Question 677579: A question: Show that the line x+16y+209=0 is normal to the curve y=10x-3x^2 at one of the points of intersection, and find this point?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Show that the line x+16y+209=0 is normal to the curve y=10x-3x^2 at one of the points of intersection, and find this point?
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y=10x-3x^2
x+16y+209=0
..
x+16(10x-3x^2)+209=0
x+160x-48x^2+209=0
48x^2-161x-209=0
solve for x by quadratic formula as follows:
a=48, b=-161, c=-209
ans:
x=-1 (At one of the points of intersection)
y=10x-3x2=-10-3=-13
point of intersection: (-1,-13)
or
x≈4.35417
..
Take derivative of first equation:y=10x-3x^2
y'=10-6x (this is an equation of slope as a function of x)
slope at point of intersection=10-6(-1)=16
..
x+16y+209=0
16y=-x-209
y=-x/16-209/16
slope=-1/16
This slope is the negative reciprocal of the slope of the curve at the point of intersection
Therefore, equation of the line,x+16y+209=0, is normal to the curve, y=10x-3x^2,at the point of intersection(-1,-13)
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