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02-13-09, 01:18 PM
Problem:
GIven a parabola -x^2 + x. Where does the line normal to this parabola at (1,0) intersect the parabola a second time?
So my thought is to first differentiate the parabola, giving me f'(x) = -2x + 1. Derivative is a formula to find hte slope of the tangent line, so plug in 1 for x: -2(1) + 1 = -1.
So we line whose slope is -1 and passes through the points (1,0), giving me a line in the form of y=-x+1
Normal line is perpendicular to the tangent line, so the equation of the normal line is y=(1/x) + 1
From here on I am not so certain what to do, how to find the second intersection point. Would a system of equations be my best bet? Set the equation of the normal line equal to the equation of the parabola and solve for x? Having an X^2, wouldn't I have two values of x, each one corresponding to a different intersection point?
GIven a parabola -x^2 + x. Where does the line normal to this parabola at (1,0) intersect the parabola a second time?
So my thought is to first differentiate the parabola, giving me f'(x) = -2x + 1. Derivative is a formula to find hte slope of the tangent line, so plug in 1 for x: -2(1) + 1 = -1.
So we line whose slope is -1 and passes through the points (1,0), giving me a line in the form of y=-x+1
Normal line is perpendicular to the tangent line, so the equation of the normal line is y=(1/x) + 1
From here on I am not so certain what to do, how to find the second intersection point. Would a system of equations be my best bet? Set the equation of the normal line equal to the equation of the parabola and solve for x? Having an X^2, wouldn't I have two values of x, each one corresponding to a different intersection point?