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

Concept: Standard form of a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ — transverse axis along the $x$-axis.

Step 1: Identify $a^2 = 16$ and $b^2 = 9$, so $a = 4$, $b = 3$.

Step 2: For a hyperbola, $c^2 = a^2 + b^2 = 16 + 9 = 25$, so $c = 5$.

Step 3: Vertices are at $(\pm a, 0) = (\pm 4, 0)$. Foci are at $(\pm c, 0) = (\pm 5, 0)$. Eccentricity $e = \frac{c}{a} = \frac{5}{4}$. Length of latus rectum $= \frac{2b^2}{a} = \frac{2 \cdot 9}{4} = \frac{9}{2}$.

The equation $\frac{x^2}{16} - \frac{y^2}{9} = 1$ is already in the standard form of a hyperbola that opens left and right. The standard form is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where the centre is at the origin. The key idea is that for this type, the transverse axis lies along the x-axis, so the vertices and foci lie on the x-axis. The relationship between $a$, $b$, and the distance to the foci $c$ is $c^2 = a^2 + b^2$ — note the plus sign, which is different from an ellipse.

Let’s extract the values directly from the denominators.

1. Identify $a$ and $b$.

   Here $a^2 = 16$, so $a = 4$. And $b^2 = 9$, so $b = 3$.

   $a$ is the distance from the centre to each vertex along the transverse axis. $b$ relates to the asymptotes and the latus rectum.

2. Find $c$ (distance from centre to each focus).

   For a hyperbola, $c^2 = a^2 + b^2$.

The foci are on the x-axis, so their coordinates are $(\pm c, 0) = (\pm 5, 0)$.

3. Vertices.

   The vertices are at $(\pm a, 0) = (\pm 4, 0)$.

4. Eccentricity $e$.

   Eccentricity for a hyperbola is defined as $e = \frac{c}{a}$.

Since $e > 1$, this confirms it’s a hyperbola.

5. Length of the latus rectum. The latus rectum of a hyperbola is a chord through a focus, perpendicular to the transverse axis. Its length is given by $\frac{2b^2}{a}$.