Conic Sections - Eccentricity (Hyperbola)

Find the eccentricity of the hyperbola given by the equation $9x^2 - 16y^2 = 144$.

First, we convert the equation into the standard form by dividing both sides by $144$: $\frac{9x^2}{144} - \frac{16y^2}{144} = \frac{144}{144}$ $\frac{x^2}{16} - \frac{y^2}{9} = 1$ Comparing this with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we get: $a^2 = 16 \implies a = 4$ $b^2 = 9 \implies b = 3$ To find $c$, we use $c^2 = a^2 + b^2$: $\begin{array}{r} 16 \\ + 9 \\ \hline 25 \end{array}$ So, $c^2 = 25 \implies c = 5$. Now, the eccentricity $e$ is: $e = \frac{c}{a} = \frac{5}{4} = 1.25$.

The equation is first normalized to identify the values of $a$ and $b$. Then, the distance to the focus $c$ is calculated using the Pythagorean-like relation for hyperbolas. Finally, the ratio $c/a$ gives the eccentricity.

Find the equation of a hyperbola with foci $(\pm 5, 0)$ and eccentricity $e = \frac{5}{4}$.

The foci are given as $(\pm c, 0) = (\pm 5, 0)$, so $c = 5$. Given eccentricity $e = \frac{5}{4}$. Using the formula $e = \frac{c}{a}$: $\frac{5}{4} = \frac{5}{a}$ By cross-multiplication, $5a = 20$, which gives $a = 4$. We know $b^2 = c^2 - a^2$. Substituting the values: $b^2 = 5^2 - 4^2$ $b^2 = 25 - 16 = 9$ The standard equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Substituting $a^2 = 16$ and $b^2 = 9$: $\frac{x^2}{16} - \frac{y^2}{9} = 1$

Since the foci are on the x-axis, we use the horizontal hyperbola form. We find $a$ from the eccentricity formula and then calculate $b^2$ using the relationship between $a, b,$ and $c$ before writing the final equation.