A Funny Thing Happened On The Way To Infinity...

Same concept, maybe simpler calculations.

$\begin {cases} 3x^2 - 12x +2y^2+12y+6 = 0 \\ xy-2y+3x = k \end {cases} \quad \Rightarrow \begin {cases} 3(x-2)^2 + 2(y+3)^2 = 24 \\ (x-2)(y+3) = k - 6 \end {cases}$

$\Rightarrow \begin {cases} 3u^2 + 2v^2 = 24 \\ uv = k - 6 \end {cases}$

$\begin {cases} \dfrac {d }{du}(3u^2 + 2v^2 - 24) = 6u + 4v\dfrac {dv}{du} \\ \dfrac {d}{du} (uv - k + 6) = v + u \dfrac {dv}{du} \end {cases} \quad \Rightarrow \begin {cases} \dfrac {dv}{du} = - \dfrac {3u}{2v} \\ \dfrac {dv}{du} = - \dfrac {v}{u} \end {cases}$

At the point of tangent: $- \dfrac {3u}{2v} = - \dfrac {v}{u} \quad \Rightarrow 3u^2 = 2v^2$

From $3u^2 + 2v^2 = 24 \quad \Rightarrow 6u^2 = 24 \quad \Rightarrow u = \pm 2 \quad \Rightarrow v = \pm \sqrt{6}$

Substituting the values of $u$ and $v$ in $uv = k - 6$ : $\Rightarrow k = 2\sqrt{6} + 6 = a\sqrt{b}+c \quad \Rightarrow a + b + c = \boxed {14}$

The equation for the ellipse can be written as

$\dfrac{(x - 2)^{2}}{8} + \dfrac{(y + 3)^{2}}{12} = 1$ ,

and the equation for the hyperbola can be written as

$(x - 2)(y + 3) = k - 6$ .

Now let $x_{0} = x - 2$ , $y_{0} = y + 3$ , $p = 2\sqrt{2}$ and $q = 2\sqrt{3}$ . Then the equations of the ellipse and hyperbola are respectively transformed into

$\dfrac{x_{0}^{2}}{p^{2}} + \dfrac{y_{0}^{2}}{q^{2}} = 1$ and $x_{0}*y_{0} = k - 6$ .

Now a necessary condition for the ellipse and hyperbola to be tangent is that they have one point of intersection for some $x_{0} \gt 0, y_{0} \gt 0$ . This point of intersection will occur when $y_{0} = \dfrac{k - 6}{x_{0}}$ . Substituting this back into the transformed ellipse equation yields that

$\dfrac{x_{0}^{2}}{p^{2}} + \dfrac{(k - 6)^{2}}{(x_{0} q)^{2}} = 1 \Longrightarrow q^{2}x_{0}^{4} - p^{2}q^{2}x_{0}^{2} + p^{2}(k - 6)^{2} = 0$ ,

which is a quadratic in $x_{0}^{2}$ . Since we want a single point of intersection (in the first quadrant of the transformed coordinates), the discriminant of this quadratic must equal $0$ , i.e.,

$p^{4}q^{4} - 4p^{2}q^{2}(k - 6)^{2} = 0 \Longrightarrow p^{2}q^{2} - 4(k - 6)^{2} = 0$

$\Longrightarrow (pq)^{2} = (2(k - 6))^{2} \Longrightarrow k - 6 = \pm \dfrac{pq}{2}$ ,

where I made use of the fact that $p, q \ne 0$ . So $k = 6 \pm 2\sqrt{6}$ , and since it is specified that $a,b,c$ are all positive we choose the positive root, giving us $a + b + c = 2 + 6 + 6 = \boxed{14}$ .

Comment: It can be confirmed using calculus that the slopes of both curves at the point of intersection is the same, (i.e., $-\frac{3}{2}$ ), when $k = 2\sqrt{6} + 6$ , implying that the point of intersection is in fact a point of tangency.

Here is a graphic depiction of the scenario for $k = 2\sqrt{6} + 6$ , and here is the graph for $k = -2\sqrt{6} + 6$ . In both cases $k \gt 0$ and both satisfy the stated requirements up until the one that specifies $a$ as being positive.