The shortest distance between point A(p,q), parabola y = root(2px)

[Pam_pam]

Hello, i have trouble finding the shortest distance between point A(p,q) and parabola y = root(2px). (p and q - sustainables)
I tried using this formula { d = root[(x1-x2)^2 + (y1-y2)^2] }
x1 = p
x2 = x
y1 = q
y2 = root(2px)
And when i had filled these variables into formula i counted derivative, however, in equation of numerator with 0 - was unreal to find (x) or i did mistake.
I would be happy if someone told me if my way of doing this task was wrong (and tell me better way) or the problem is that i did mistake.
I hope i created this post in appropriate place.

 

[ksdhart2]

The steps you've outlined seem reasonable to me, but, unfortunately, without actually seeing your work, we can't critique it and tell you where you may or may not have gone wrong. Please share all of your work with us, even if you're sure it's wrong. Thank you.

 

[HallsofIvy]

By the way, finding a max or min for \(\displaystyle \sqrt{f(x)}= (f(x))^{1/2}\) involves finding x such that the derivative, \(\displaystyle \frac{1}{2}f(x)^{-1/2} f'(x)\), is 0. Since a fraction is 0 if and only if its numerator is 0, that is, if and only if f'(x) is 0.

In other words, in problems that ask for a minimum distance, it is sufficient, and simpler, to minimize distance squared. Here, that would be \(\displaystyle (x- p)^2+ (2\sqrt{px}- q)^2= x^2+ 4px- 2px+ p^2+ q^2\)\(\displaystyle -4q*\sqrt{px}\) .. edited

By the way, are the "p" in (p, q) and \(\displaystyle 2\sqrt{px}\) the same? That seems peculiar.

 

[Steven G]

Since a fraction is 0 if and only if its numerator is 0

 

[HallsofIvy]

\(\displaystyle \frac{a}{b}= 0\) if and only if a= 0. Surely that does not surprise you?