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?
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 $x=-\frac{1}{34}$.
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 $$ \sqrt{L(-1/34)} = \sqrt{\frac{1359}{68}} \approx 4.47. $$
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.
As per this post, the shortest distance between $2$ curves is along their common normal. Now, consider the curves $y=\sqrt{16x^2+5x+16}$ and $(x-2)^2+y^2=r^2$ where $r$ is some real constant. It is clear that any common normal between these $2$ curves will pass through $(2,0)$ as the second curve represents a circle with this point as its centre. On taking the limiting situation of $r\to0$, the second curve tends to the point $(2,0)$ but any common normal will still pass through this point.
Hence, the shortest distance between $f(x)=\sqrt{16x^2+5x+16}$ and $(2,0)$ will occur when the normal to $f$ passes through $(2,0)$. Let this normal be drawn to $f$ at the point $(\alpha,\beta)$. Then,
$$\cancel\beta=-(\alpha-2)\frac{2\cancel{\sqrt{16\alpha^2+5\alpha+16}}}{32\alpha+5}$$ $$\implies32\alpha+5=4-2\alpha$$ $$\implies\alpha=\frac{-1}{34}$$
Now, all that remains is some tedious algebraic calculations.
Comment
In the meanwhile if OP has picked up interest in advanced methods in Variational calculus (Euler Lagrange equations related)
$$ F=16x^2+5x +16 -y^2,~~ G= x^2+y^2;~$$
When using partial derivatives
$$\frac{F_x}{F_y}=\frac{G_x}{G_y}$$
$$\frac{32 x+5}{-2y}=\frac{2x-4}{2y}~\to ~ x=\frac{-1}{34}$$
and $y$,$ \sqrt{x^2+y^2}$ can be found, as also the minimum distance to given hyperbola from the origin.
