Q.Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the hyperbolas:

Concept: Standard form of a hyperbola and its parameters.

For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (horizontal transverse axis), we have $c^2 = a^2 + b^2$, vertices at $(\pm a, 0)$, foci at $(\pm c, 0)$, eccentricity $e = \frac{c}{a}$, and latus rectum $\frac{2b^2}{a}$.

(i) $\frac{x^2}{9} - \frac{y^2}{16} = 1$

Here $a^2 = 9$, $b^2 = 16$, so $a = 3$, $b = 4$, and $c = \sqrt{9+16} = 5$.

• Vertices: $(\pm 3, 0)$
• Foci: $(\pm 5, 0)$
• Eccentricity: $e = \frac{5}{3}$
• Latus rectum: $\frac{2 \times 16}{3} = \frac{32}{3}$

(ii) $y^2 - 16x^2 = 16 \implies \frac{y^2}{16} - \frac{x^2}{1} = 1$

This has a vertical transverse axis with $a^2 = 16$, $b^2 = 1$, so $a = 4$, $b = 1$, and $c = \sqrt{16+1} = \sqrt{17}$.

• Vertices: $(0, \pm 4)$
• Foci: $(0, \pm \sqrt{17})$
• Eccentricity: $e = \frac{\sqrt{17}}{4}$
• Latus rectum: $\frac{2 \times 1}{4} = \frac{1}{2}$

A hyperbola is the locus of points where the absolute difference of distances to two fixed points (the foci) is constant. The standard forms tell us immediately which axis the hyperbola straddles:

The positive term dictates the transverse axis. The relationship $c^2 = a^2 + b^2$ (note the plus sign, unlike ellipses) locates the foci at distance $c$ from the center along the transverse axis. The vertices sit at $\pm a$ on that axis, the eccentricity is $e = \frac{c}{a} > 1$, and the latus rectum—the chord through a focus perpendicular to the transverse axis—has length $\frac{2b^2}{a}$.

(i) $\frac{x^2}{9} - \frac{y^2}{16} = 1$

1. Identify the form and parameters.

   The equation is already in standard form with the $x^2$ term positive, so this is a horizontal hyperbola centered at the origin. Reading off: $a^2 = 9 \implies a = 3$ and $b^2 = 16 \implies b = 4$.

2. Compute $c$.

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

3. Vertices.

   The transverse axis is horizontal, so the vertices lie at $(\pm a, 0)$:

4. Foci. The foci are at $(\pm c, 0)$:

5. Eccentricity.

6. Length of the latus rectum.

(ii) $y^2 - 16x^2 = 16$

1. Rewrite in standard form. Divide through by $16$:

Now the $y^2$ term is positive, so this is a vertical hyperbola. We have $a^2 = 16 \implies a = 4$ (on the $y$-axis) and $b^2 = 1 \implies b = 1$ (on the $x$-axis).

2. Compute $c$.

3. Vertices. The transverse axis is vertical, so vertices are at $(0, \pm a)$:

4. Foci. The foci lie at $(0, \pm c)$:

5. Eccentricity.

6. Length of the latus rectum.