Shortest distance from point to curve I could use some help solving the following problem. I have many more like this but I figured if I learn how to do one then I can figure out the rest on my own. Thanks in advance!

A curve described by the equation $y=\sqrt{16x^2+5x+16}$ on a Cartesian plane. What is the shortest distance between coordinate $(2,0)$ and this line?

 A: Start by finding the distance from some point on the curve to $(2,0)$ in terms of $x$. Using the distance formula, we get
$$D=\sqrt{(x-2)^2+(\sqrt{16x^2+5x+16}-0)^2}$$
$$D=\sqrt{x^2-4x+4+16x^2+5x+16}$$
$$D=\sqrt{17x^2+x+20}$$
This will end up being a messy derivative. However, since the distance $D$ will never be negative, we can minimize $D^2$ instead of $D$ and still get the same answer. So now we get
$$D^2=17x^2+x+20$$
$$\frac{dD^2}{dx}=34x+1$$
Now we set this equal to $0$ and solve for $x$:
$$34x+1=0$$
$$x=-\frac{1}{34}$$
So the distance is minimized at $x=-\frac{1}{34}$, and to find the minimum distance, simply evaluate $D$ when $x=-\frac{1}{34}$.
A: Since distance is positive and the square root function is increasing, it suffices to find the smallest value the squared distance between $(x,y)$ on the curve and the point $(2,0)$ can take. This is
$$ L(x) = (x-2)^2 + (y-0)^2 = (x-2)^2+y^2 = x^2-4x+4 + 16x^2+5x+16 = 17x^2+x+20. $$
A minimum can only occur if $L'(x)=0$. So
$$ L'(x) = 34x+1, $$
so there is a turning point at $x=-1/34$. Moreover, the derivative is negative on the left and positive on the right, so the point is a minimum. Hence the minimum distance is
$$ \sqrt{L(-1/34)} = \sqrt{\frac{1359}{68}} \approx 4.47. $$
A: Hint 1: take a point $(x, y)$ on the curve, calculate its distance from $(2, 0)$. The fact that the point is on the curve allows you to express that distance in terms of $x$ alone. Then find the minimum (but check the second hint first).
Hint 2: instead of minimizing the distance, minimize the square of the distance.
