Ellipse Sandwich

The whole figure is symmetric about the bisector of the angle formed by the two lines, so we can position it on a coordinate system with the center of the ellipse being the origin, the major axis of the ellipse on the x-axis and the centers of the circles on the y-axis, as shown in the diagram. Let the equation of the ellipse be $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ , where $a$ and $b$ are its major and minor semi-axis respectively.

The centers of the small and the big circles are $K\left( 0,b+4 \right)$ and $L\left( 0,-b-9 \right)$ respectively.

The equation of the line $l$ that intersects the positive coordinate semi-axes $Ox$ and $Oy$ is of the form $y=mx+c\Leftrightarrow mx-y+c=0$ where the y-intercept $c$ is greater than the y-coordinate of $K$ , i.e. $c>b+4>0$ .

Since $l$ is tangent to the circles, the distances from their centers to the line equal their radii. Using the formula for the Distance between Point and a Line we have

$d\left( K,l \right)=\dfrac{\left| -b-4+c \right|}{\sqrt{{{m}^{2}}+1}}=4\Leftrightarrow -b-4+c=4\sqrt{{{m}^{2}}+1}\Leftrightarrow c-b=4\sqrt{{{m}^{2}}+1} +4 \ \ \ \ \ \left( 1 \right)$

$d\left( L,l \right)=\frac{\left| b+9+c \right|}{\sqrt{{{m}^{2}}+1}}=9\Leftrightarrow c+b=9\sqrt{{{m}^{2}}+1}-9\ \ \ \ \ \ \left( 2 \right)$

Line $l$ is also tangent to the ellipse, i.e. the line and the ellipse have exactly one common point. This means that the system of their equations has exactly one solution. Combining the two equations we get

$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{\left( mx+c \right)}^{2}}}{{{b}^{2}}}=1\Leftrightarrow \left( {{a}^{2}}{{m}^{2}}+{{b}^{2}} \right){{x}^{2}}+\left( 2mc{{a}^{2}} \right)x+\left( {{a}^{2}}{{c}^{2}}-{{a}^{2}}{{b}^{2}} \right)=0$ The latter is a quadratic equation and, since it has a unique solution, the discriminant is 0, i.e.

${{\left( 2mc{{a}^{2}} \right)}^{2}}-4\left( {{a}^{2}}{{m}^{2}}+{{b}^{2}} \right)\left( {{a}^{2}}{{c}^{2}}-{{a}^{2}}{{b}^{2}} \right)=0$ Rearranging,

$\begin{aligned} {{a}^{2}}{{m}^{2}}={{c}^{2}}-{{b}^{2}} & \Rightarrow {{a}^{2}}{{m}^{2}}=\left( c-b \right)\left( c+b \right) \\ & \overset{\left( 1 \right),\left( 2 \right)}{\mathop{\Rightarrow }}\,{{a}^{2}}{{m}^{2}}=\left( 4\sqrt{{{m}^{2}}+1}+4 \right)\left( 9\sqrt{{{m}^{2}}+1}-9 \right) \\ & \Rightarrow {{a}^{2}}{{m}^{2}}=36{{m}^{2}} \\ & \overset{m\ne 0}{\mathop{\Rightarrow }}\,a=\boxed{6} \\ \end{aligned}$

Rotate the diagram so that the angle bisector is on the $y$ -axis and so that the origin is the center of the ellipse:

Let the semi-major axis of the ellipse be $a$ and the semi-minor axis be $b$ . Then the equation of the ellipse is $\cfrac{x^2}{a^2} + \cfrac{y^2}{b^2} = 1$ , the equation of the top circle is $x^2 + (y - b - 4)^2 = 16$ , and the equation of the bottom circle is $x^2 + (y + b + 9)^2 = 81$ .

Let the equation of the tangent line be $y = mx + c$ . Then the top circle will only have one $x$ -value for $x^2 + (mx + c - b - 4)^2 = 16$ , which means that its discriminant will be zero, or $(c - b)^2 - 8(c - b) - 16m^2 = 0$ , and this solves to $c - b = 4(\sqrt{m^2 + 1} + 1)$ .

Similarly, the bottom circle will only have one $x$ -value for $x^2 + (mx + c + b + 9)^2 = 81$ , which means that its discriminant will be zero, or $(c + b)^2 - 18(c + b) - 81m^2 = 0$ , and this solves to $c + b = 9(\sqrt{m^2 + 1} - 1)$ .

Similarly, the ellipse will only have one $x$ -value for $\cfrac{x^2}{a^2} + \cfrac{(mx + c)^2}{b^2} = 1$ , which means that its discriminant will be zero, or $a^2m^2 + b^2 - c^2 = 0$ , and this solves to $a^2 = \cfrac{(c - b)(c + b)}{m^2}$ .

Substituting $c - b = 4(\sqrt{m^2 + 1} + 1)$ and $c + b = 9(\sqrt{m^2 + 1} - 1)$ into $a^2 = \cfrac{(c - b)(c + b)}{m^2}$ gives $a^2 = \cfrac{4(\sqrt{m^2 + 1} + 1) \cdot 9(\sqrt{m^2 + 1} - 1)}{m^2} = 36$ , which solves to $a = 6$ .

Therefore, the semi-major axis of the ellipse is $a = \boxed{6}$ .