Conic Sections - Standard equation of Hyperbola

Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum for the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Comparing with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, we get $a^2 = 16$ and $b^2 = 9$. Therefore, $a = 4$ and $b = 3$. We find $c = \sqrt{a^2 + b^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

1. Vertices: $(\pm a, 0) = (\pm 4, 0)$
2. Foci: $(\pm c, 0) = (\pm 5, 0)$
3. Eccentricity: $e = \frac{c}{a} = \frac{5}{4}$
4. Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{18}{4} = 4.5$

Identify the orientation (horizontal since $x^2$ is positive), extract $a$ and $b$, calculate $c$ using the hyperbola identity, and apply standard coordinate formulae.

Find the equation of the hyperbola with foci $(0, \pm 13)$ and conjugate axis of length $24$.

Since the foci are on the $y$-axis, the transverse axis is along the $y$-axis. The general equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. Given foci $(0, \pm c) = (0, \pm 13) \implies c = 13$. Given length of conjugate axis $2b = 24 \implies b = 12$. Using $c^2 = a^2 + b^2$, we have: $13^2 = a^2 + 12^2$ $169 = a^2 + 144$ $a^2 = 169 - 144 = 25$ Therefore, the equation is: $\frac{y^2}{25} - \frac{x^2}{144} = 1$

Determine the orientation from the foci. Use the given conjugate axis length to find $b$. Solve for $a^2$ using the relationship between $a, b, c$, then substitute into the vertical hyperbola standard form.