Always Tangential

If $y(x) = x^2 - 2ax + a^2 + 14$ for all $a \in \mathbb{R}$ , then $m = \frac{dy}{dx}|_{x=x_{0}} = 2x_{0}-2a$ . The tangent line at $(x_{0}, y(x_{0}))$ can be written as:

$y - (x_{0}^2 - 2ax_{0} + a^2 + 14) = (2x_{0}-2a)(x-x_{0})$ ;

or $y = (2x_{0} - 2a)x + (2ax_{0} - 2x_{0}^{2} + x_{0}^2 - 2ax_{0} + a^2 + 14);$

or $y = (2x_{0} - 2a)x + (-x_{0}^{2} + a^2 + 14) = mx+n.$

Taking $m+n$ gives us:

$(2x_{0} - 2a) + (-x_{0}^{2} + a^2 + 14) = (a-1)^2 - (x_{0}-1)^2 + 14$

and for all $x_{0} = a$ , we obtain $\boxed{m+n = 14}$ .