# Thread: minimise/maximise distance between (0,4) and y=x^2/4

1. ## minimise/maximise distance between (0,4) and y=x^2/4

I am trying to find the minimum and maximum distance between point (0,4) and the line y=(x^2)/4 in the region of 0<=x<=2√3.

This is what i have so far

y(distance)=√((x-0)^2+((x^2)/4)^2)
=(x^2+(X^4)/16+16)^0.5
then calculate y'
then make y'=0
then determine if max or min using 2nd derivative test

The problem i am having is that the answers i get for x when y' = 0 are imaginary. i have computed this problem using matlab and get the same answers except that they are real and have isolated the problem to my simplification of y.

matlab gives it as y=0.25(-16x^2+x^4+256)^0.5

Could some one please explain how to simplify y=(x^2+(X^4)/16+16)^0.5 to y=0.25(-16x^2+x^4+256)^0.5 or suggest a better way to approach this problem

2. Originally Posted by Poppy
I am trying to find the minimum and maximum distance between point (0,4) and the line y=(x^2)/4 in the region of 0<=x<=2√3.

This is what i have so far

y(distance)=√((x-0)^2+((x^2)/4)^2)
=(x^2+(X^4)/16+16)^0.5
then calculate y'
then make y'=0
then determine if max or min using 2nd derivative test

The problem i am having is that the answers i get for x when y' = 0 are imaginary. i have computed this problem using matlab and get the same answers except that they are real and have isolated the problem to my simplification of y.

matlab gives it as y=0.25(-16x^2+x^4+256)^0.5

Could some one please explain how to simplify y=(x^2+(X^4)/16+16)^0.5 to y=0.25(-16x^2+x^4+256)^0.5 or suggest a better way to approach this problem

Remember that the distance between two points $\displaystyle (x_1, y_1)$ and $\displaystyle (x_2, y_2)$ is given by

$\displaystyle d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

(this is easily proven by Pythagoras).

In your case $\displaystyle (x_1, y_1) = (0, 4)$ and $\displaystyle (x_2, y_2) = \left(x, \frac{x^2}{4}\right)$.

So the distance is

$\displaystyle d = \sqrt{(x - 0)^2 + \left(\frac{x^2}{4} - 4\right)^2}$

$\displaystyle = \sqrt{x^2 + \left(\frac{x^2 - 16}{4}\right)^2}$

$\displaystyle = \sqrt{x^2 + \frac{(x^2 - 16)^2}{16}}$

$\displaystyle = \sqrt{\frac{16x^2 + (x^2 - 16)^2}{16}}$

$\displaystyle = \frac{\sqrt{16x^2 + x^4 - 32x + 256}}{4}$

$\displaystyle = \frac{\sqrt{x^4 + 16x^2 - 32x + 256}}{4}$.

Now you can find the minimum and maximum...

3. ## Thanks-just one error

thanks for that it was very helpful- only one minor correction -when you expanded (x^2-16)^2 it should have been to x^4-32x+256 not x^2-32x+256 which makes the next equation ((16x^2^2+x^4-32x+256)/4)^1/2 not ((17x^2-32x+256)/4)^1/2

4. Edited, thanks. It's well known that my arithmetic is shocking...

However, to correct YOUR latest post, it actually becomes

$\displaystyle \frac{(x^4 + 16x^2 - 32x + 256)^{\frac{1}{2}}}{4}$

not

$\displaystyle \left(\frac{x^4 + 16x^2 - 32x + 256}{4}\right)^\frac{1}{2}$.