Conic Sections - Relationship between semi-major axis, semi-minor axis and the distance of the focus

Given the equation of an ellipse $\frac{x^2}{169} + \frac{y^2}{144} = 1$, find the distance of the focus from the center and the eccentricity.

1. Identify $a^2$ and $b^2$: Here $a^2 = 169$ and $b^2 = 144$.
2. Calculate $a$ and $b$: $a = \sqrt{169} = 13$, $b = \sqrt{144} = 12$.
3. Find $c$ using $c^2 = a^2 - b^2$: $\begin{array}{r} 169 \\ - 144 \\ \hline 25 \end{array}$ $c^2 = 25 \implies c = 5$.
4. Calculate eccentricity $e$: $e = \frac{c}{a} = \frac{5}{13}$.

In an ellipse, $a$ is the semi-major axis. We use the subtraction relationship to find the distance of the focus ($c$). The eccentricity is then the ratio of $c$ to $a$.

Find the distance of the focus ($c$) for a hyperbola with semi-transverse axis $a = 6$ and semi-conjugate axis $b = 8$.

1. Given $a = 6$ and $b = 8$.
2. Use the hyperbola relationship $c^2 = a^2 + b^2$.
3. Calculate squares: $a^2 = 36$, $b^2 = 64$.
4. Sum the values: $\begin{array}{r} 36 \\ + 64 \\ \hline 100 \end{array}$
5. Find $c$: $c = \sqrt{100} = 10$.

For a hyperbola, the distance to the focus is always greater than the semi-axes, following the Pythagorean relationship $c^2 = a^2 + b^2$.