# Thread: closes points to points on the curve

1. ## closes points to points on the curve

I have this question:
which point on the graph of y = 4 - x^2 are the closest to the point (0,2)?
i know the graph is going to be a downward parabola mirrored across the y axis started at the point 4, the point (o,2) is inside the parabola.
For this problem do I just take the derivative and then evaluate the the x value of zero ? Can anyone explain the concept of this problem ?
I know the derivative will give me the tangent line to the curve, Then whats my next step?
-Marcus

2. ## Re: closes points to points on the curve

The quadratic distance of a point of $y=4-x^{2}$ and the point (0,2) is...

$d^{2}= (y-2)^{2} + (x-0)^{2} = (2-x^{2})^{2} + x^{2}$ (1)

... then find the x that mininizes (1)...

Kind regards

$\chi$ $\sigma$

3. ## Re: closes points to points on the curve

You can prove that:

$|min\{(2-x^{2})^{2} + x^{2} \}|=t \ \tex{when} \ \ x=\sqrt{\frac{3}{2}}$

4. ## Re: closes points to points on the curve

I think i need to solve this equation using calculus. Any other suggestions?

5. ## Re: closes points to points on the curve

Originally Posted by mjfuentes85
I think i need to solve this equation using calculus. Any other suggestions?
4-4x^2+x^4+x^2=x^4-3x^2+4=f(x)

x^2=t

t^2-3t+4=g(t)

no roots. min at: t=3/2

t=x^2=3/2 ===> x=sqrt{3/2}

6. ## Re: closes points to points on the curve

Use what chisigma told you. You have a formula for the distance between any point on the curve with the coordinates (0,2).

How can you minimise that equation? By using calculus, by finding the derivative of the equation and setting it to 0 to get the point x on the curve that is closest to (0,2), which you can then use to get the y-coordinate of that point.

7. ## Re: closes points to points on the curve

Originally Posted by mjfuentes85
I have this question:
which point on the graph of y = 4 - x^2 are the closest to the point (0,2)?
i know the graph is going to be a downward parabola mirrored across the y axis started at the point 4, the point (o,2) is inside the parabola.
For this problem do I just take the derivative and then evaluate the the x value of zero ? Can anyone explain the concept of this problem ?
I know the derivative will give me the tangent line to the curve, Then whats my next step?
-Marcus
1. The shortest distance between P(0, 2) and a point on the curve is the perpendicular line segment to the point T(t, 4-tē). Obviously T lies on the parabola.

2. The slope $m_t$ of the tangent in T is

$y'=-2x~\implies~m_t=-2t~\implies~m_n = \dfrac1{2t}$

where $m_n$ is the slope of the normal (= perpendicular line) to the tangent.

3. The equation of the normal is:

$n: y-(4-t^2)=\dfrac1{2t}(x-t)$

4. The point P belongs to the normal. Plug in it's coordinates and solve the equation for t:

$2-(4-t^2)=\dfrac1{2t}(-t)~\implies~t=\pm\sqrt{\dfrac32}$

5. You'll get 2 different points T. Calculate the distance $|\overline{PT}|$. You should come out with $\frac12 \sqrt{7}$