Distance On The Cartesian Plane

This solution involves Calculus .

Consider the restriction $g(x,y) = x^2-y-4 = 0$ and the point $P = (x_0,y_0)$ such that $g(P) = 0$ . We want to minimize $D = d((0,2),P) = \sqrt{(x_0-0)^2 + (y_0-2)^2} \Rightarrow [d((0,2),P)]^2 = x_0^2 + (y_0-2)^2 = f(P)$ .

See that $\vec{\bigtriangledown}f(P) = (2x_0, 2y_0-4), \; \vec{\bigtriangledown}g(P) = (2x_0,-1) \neq \vec{0}, \; \forall P$ , and since both function $f,g$ are of class $C^1$ (first derivatives are continuous), we can apply the method of Lagrange multipliers :

$\vec{\bigtriangledown}f(P) + \lambda \vec{\bigtriangledown}g(P) = \vec{0} \Rightarrow \left\{\begin{matrix} x_0(\lambda +1) =0 \Rightarrow \left\{\begin{matrix} x_0 = 0\\ \lambda = -1 \end{matrix}\right. \\ y_0 = 2 + \frac{\lambda}{2} \end{matrix}\right.$

For $x_0 = 0$ :

$g(0,y_0) = 0 \Rightarrow y_0 = -4$ . Thus $D = 6$ .

For $\lambda = -1$ :

$y_0 = \frac{3}{2}$ . For $g\left ( x_0,\frac{3}{2} \right ) = 0 \Rightarrow x_0 = \pm \sqrt{\frac{11}{2}}$ . Thus $D = \sqrt{\frac{23}{4}} < 6$ .

Finally, $D^2 = \frac{23}{4} = \boxed{5.75}$ .