### Instruction: Point $O$ is the center of an ellipse with major axis $\overline{AB}$ and minor axis $\overline{CD}.$  Point $F$ is one focus of the ellipse.  If $OF = 6$ and the diameter of the inscribed circle of triangle $OCF$ is 2, compute the product $(AB)(CD).$.### Response: I know that the distance from the center of an ellipse to each focus is $c$, where $a^2 = b^2 + c^2,$ where $a$ and $b$ are the semi-major and semi-minor axes, respectively.
So, in this problem, $c = 6.$
I also know that the diameter of the incircle of a right triangle is equal to the sum of the two legs minus the hypotenuse.
So, in this problem, $2 = OC + CF - OF.$
Substituting $OF = 6,$ I get $OC + CF = 8.$
But, I also know that $OC = b$ and $CF = a - c.$
So, I can rewrite this equation as $b + a - c = 8.$
Substituting $c = 6,$ I get $b + a - 6 = 8.$
Solving for $a,$ I get $a = 14 - b.$
Now, I have two equations relating $a, b,$ and $c: a^2 = b^2 + c^2$ and $a = 14 - b.$
I can substitute the second equation into the first and get $(14 - b)^2 = b^2 + 6^2.$
Expanding and simplifying, I get $b^2 - 28b + 196 = b^2 + 36.$
Subtracting $b^2$ from both sides, I get $-28b + 160 = 0.$
Adding $28b$ to both sides, I get $160 = 28b.$
Dividing both sides by 28, I get $b = \frac{160}{28} = \frac{40}{7}.$
Now, I can plug this value of $b$ into either equation to find $a.$
I choose the simpler one: $a = 14 - b.$
So, $a = 14 - \frac{40}{7} = \frac{98 - 40}{7} = \frac{58}{7}.$
Therefore, the product of the major and minor axes is $(2a)(2b) = 4ab.$
Plugging in the values of $a$ and $b,$ I get $4ab = 4(\frac{58}{7})(\frac{40}{7}) = \frac{9280}{49}.$