Calculus, Closest Point on a Parabola
- Thread starter chris2005
- Start date
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.
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.
Hello, chris2005!
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.
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.