Distance Between the Foci

1. Identify the given relations for the hyperbola
For the standard hyperbola $$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ with eccentricity $$e$$:
The distance between the two foci is given as $$2ae = 20 \implies ae = 10$$
The distance between the two directrices is given as $$\frac{2a}{e} = 10 \implies \frac{a}{e} = 5$$

2. Calculate the values of $$a^2 and e^2$$
Multiply the two simplified equations to eliminate $$e$$:
$$(ae) \times \left(\frac{a}{e}\right) = 10 \times 5$$
$$a^2 = 50$$

Divide the first equation by the second equation to eliminate $$a$$:
$$\frac{ae}{\left(\frac{a}{e}\right)} = \frac{10}{5}$$
$$e^2 = 2$$

3. Determine the value of $$b^2$$
For a hyperbola, the relationship between the semi-axes and eccentricity is defined by the formula:
$$b^2 = a^2(e^2 - 1)$$

Substitute $$a^2 = 50$$ and $$e^2 = 2$$ into the relation:
$$b^2 = 50(2 - 1)$$
$$b^2 = 50(1) = 50$$

4. Compute the value of k
Since $$a^2 = 50$$ and $$b^2 = 50$$, taking the positive square root for the lengths of the semi-axes gives:
$$a = \sqrt{50} = 5\sqrt{2}$$
$$b = \sqrt{50} = 5\sqrt{2}$$

Multiply $$a$$ and $$b$$ together to find the value of $$k$$:
$$k = ab = (5\sqrt{2}) \times (5\sqrt{2}) = 25 \times 2 = 50$$