Shortest distance from point to curve

Solution 1:

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 $x=-\frac{1}{34}$.

Solution 2:

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 $$ \sqrt{L(-1/34)} = \sqrt{\frac{1359}{68}} \approx 4.47. $$