Calculus, Closest Point on a Parabola

[chris2005]

heres the question: Find the point on the parabola 2y=x^2 that is closest to the point (-4, 1).

 

[chris2005]

its a tricky question

 

[tkhunny]

its a tricky question

No, it's just a distance formula. What's the distance of ANY point from (-4,1)?

 

[Gene]

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.

 

[soroban]

Hello, chris2005!

Find the point on the parabola $\displaystyle 2y\,=\,x^2$ that is closest to the point $\displaystyle A(-4,1)$.

As tkhunny pointed out, it's a Distance Formula problem . . . in Calculus.

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

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

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