Find the closest point on a graph to coordinates $(2;0.5)$

Question

My English isn't the best, but hopefully you will understand my question!

The question is as followed:

Find the point on $y=x^2$ that is closest to $A=(2;0.5)$

I'm gratefull for any advice, tips or answers. Thank you!

Answer 1

The equation of normal line to $y=x^2$ at the point $x_0$ is: $$y=x_0^2-\frac{1}{2x_0}(x-x_0)$$

The normal line must pass through the point $(2,0.5)$, so:

$$0.5=x_0^2-\frac{1}{2x_0}(2-x_0) \Rightarrow x_0=1\Rightarrow y_0=1.$$

Finally, the distance between the points $(2,0.5)$ and $(1,1)$ is:

$$d=\sqrt{(2-1)^2+(0.5-1)^2}=\frac{\sqrt{5}}{2}.$$

Answer 2

The distance from $(x,x^2)$ to $(2,\frac12)$ is $\sqrt{(x-2)^2+(x^2-\frac12)^2}$. Hence it suffices to minimize the square of the distance which is $$(x-2)^2+(x^2-\frac12)^2 = x^2-4x+4+x^4-x^2+\frac14 = x^4-4x+\frac{17}{4}.$$

To do this we can take the derivative, $4x^3-4$, and set it to zero, which yields $x^3-1 = 0$ so $x = 1$. Indeed this is a minimum because the second derivative, $12x^2$, is positive at $x = 1$.

Plugging in $x = 1$ into the original distance formula we obtain $\frac{\sqrt{5}}{2}$.

Answer 3

the distance is given by $$\sqrt{(x-2)^2+(y-1/2)^2}$$ where $$y=x^2$$ and you can differentiate this with respect to $x$