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I have differential calculus problem where I must find the coordinates of a point C on a parabola. There is a point given which is D(8,4) and D is on a line that is intersecting with the parabola at C. Also, the line where there is the point D is perpendicular to the tangent of C.


This is the equation of the parabola:

f(x) = -x^2 + 4x + 2

This is coordinates of the single point given:

D (8,4) 

What are the coordinates of C?


I am having trouble with the fact that there is only point given for the intersecting line.

Thank you for helping out!

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why is this tagged graph theory? :x) –  anonymous Oct 24 '10 at 17:21

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Using your differential calculus, you should be able to write the equation for the line tangent to the parabola at a given point on the parabola. So if $(x,y)$ satisfies $y=-x^2+4x+2$ what is the equation of the tangent line? Then you should be able to find the equation for the line perpendicular to the tangent at the point on the parabola. One of these perpendiculars will pass through $(8,4)$.

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The equation of the tangent line would be y = f(a) + f'(a) * (x-a). –  Alerty Oct 24 '10 at 17:53
    
Yes, that is right. And f'(a)=-2a+4. So now you need a perpendicular to that line at the point of tangency. You should have an equation that gives the slope of a perpendicular and you have a point. –  Ross Millikan Oct 24 '10 at 18:00
    
So, y = -2(a-2)x + a^2 + 2 and to be perpendicular it would have to be y = x /( 2(a-2) ) + a^2 + 2 + b? –  Alerty Oct 24 '10 at 18:07
    
If that is right, I need to find the value of a and b. –  Alerty Oct 24 '10 at 18:12
    
Not quite. The x has to be in the numerator to be a straight line. So y=x/2/(a-2)+c where c is chosen to match the parabola at a. This gives -a^2+4a+2=a/2/(a-2)+c. If I did my algebra right, that gives c=(-2a^3+10a^2+11a+4)/2/(a-2), but you could check. So now you have a line that has to pass through (8,4). Plugging in 8 for x and 4 for y gives a single equation in a. –  Ross Millikan Oct 24 '10 at 18:19

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