Calculus, Closest Point on a Parabola

chris2005

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Nov 22, 2005
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heres the question: Find the point on the parabola 2y=x^2 that is closest to the point (-4, 1).
 
y=.5x^2
The closest point is when the line from (-4,1) is perpendicular to the parabola.
solve for m = dy/dx
The line from (-4,1) perpendicular to m is
(y-1)=(-1/m)(x+4) or
y = (-1/m)(x+4)+1
Equate the y's and you will get the X where the line meets the parabola at (X,.5X^2).

I took that route before I got TKH's short cut. Take the derivitive of the distance formula = 0 and solve. Much simpler than the way I went and it gives the same answer. :oops:
 
Hello, chris2005!

Find the point on the parabola 2y=x2\displaystyle 2y\,=\,x^2 that is closest to the point A(4,1)\displaystyle A(-4,1).
As tkhunny pointed out, it's a Distance Formula problem . . . in Calculus.

Let P\displaystyle P be any point on the parabola . . . Then we have: .P(x,x22)\displaystyle P\left(x,\,\frac{x^2}{2}\right)

The distance PA\displaystyle PA is: .D=[x(4)]2+[x221]2\displaystyle D\:=\:\sqrt{\left[x-(-4)\right]^2\,+\, \left[\frac{x^2}{2}- 1\right]^2}

And we have the function: .\(\displaystyle D\:=\:\left[\frac{x^4}{4}\.+\,8x\,+\,17\right]^{\frac{1}{2}}\) to minimize.
 
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