Question:

Step 1: Define the geometric properties of the ellipse and triangle $OCF$. Let the semi-major axis be $a$ and the semi-minor axis be $b$. The distance from the center $O$ to a focus $F$ is $c$. Given $OF = 6$, we have $c = 6$. Point $C$ is an endpoint of the minor axis, so $OC = b$. Triangle $OCF$ is a right-angled triangle with the right angle at $O$. The vertices are $O(0,0)$, $C(0,b)$, and $F(c,0)$. Step 2: Express the side lengths of triangle $OCF$. The side lengths of $\triangle OCF$ are: $OC = b$ $OF = c = 6$ $CF = \sqrt{OC^2 + OF^2} = \sqrt{b^2 + c^2} = \sqrt{b^2 + 6^2} = \sqrt{b^2 + 36}$. Step 3: Calculate the area and semiperimeter of triangle $OCF$. The area of $\triangle OCF$ is $A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times OF \times OC = \frac{1}{2} \times 6 \times b = 3b$. The diameter of the inscribed circle is $2$, so the inradius is $r = 1$. The semiperimeter $s$ of $\triangle OCF$ is $s = \frac{OC + OF + CF}{2} = \frac{b + 6 + \sqrt{b^2 + 36}}{2}$. Step 4: Apply the inradius formula to find the value of $b$. The inradius formula for a triangle is $r = \frac{A}{s}$. Substituting the values: $1 = \frac{3b}{\frac{b + 6 + \sqrt{b^2 + 36}}{2}}$ $1 = \frac{6b}{b + 6 + \sqrt{b^2 + 36}}$ $b + 6 + \sqrt{b^2 + 36} = 6b$ $\sqrt{b^2 + 36} = 5b - 6$ Square both sides: $b^2 + 36 = (5b - 6)^2$ $b^2 + 36 = 25b^2 - 60b + 36$ $0 = 24b^2 - 60b$ $0 = 12b(2b - 5)$ Since $b$ must be a positive length, $b \neq 0$. Therefore, $2b - 5 = 0 \implies b = \frac{5}{2}$. Step 5: Determine the value of $a$. For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$. We have $c = 6$ and $b = \frac{5}{2}$. $6^2 = a^2 - \left(\frac{5}{2}\right)^2$ $36 = a^2 - \frac{25}{4}$ $a^2 = 36 + \frac{25}{4} = \frac{144 + 25}{4} = \frac{169}{4}$ $a = \sqrt{\frac{169}{4}} = \frac{13}{2}$. Step 6: Calculate the product $(AB)(CD)$. The length of the major axis is $AB = 2a$. The length of the minor axis is $CD = 2b$. The product $(AB)(CD)$ is: $(AB)(CD) = (2a)(2b) = 4ab$ Substitute the values of $a$ and $b$: $4 \times \frac{13}{2} \times \frac{5}{2} = 4 \times \frac{65}{4} = 65$. The product $(AB)(CD)$ is $65$.