Q.Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $\frac{y^2}{9} - \frac{x^2}{27} = 1$.

Concept: Hyperbola in standard vertical form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

Here $a^2 = 9$ and $b^2 = 27$, so $a = 3$, $b = 3\sqrt{3}$. For a hyperbola, $c^2 = a^2 + b^2 = 9 + 27 = 36$, hence $c = 6$.

Step 1 — Vertices: For a vertical hyperbola, vertices are at $(0, \pm a) = (0, \pm 3)$.

Step 2 — Foci: Foci are at $(0, \pm c) = (0, \pm 6)$.

Step 3 — Eccentricity: $e = \frac{c}{a} = \frac{6}{3} = 2$.

Step 4 — Latus rectum length: $\frac{2b^2}{a} = \frac{2 \times 27}{3} = 18$.

The equation given is $\frac{y^2}{9} - \frac{x^2}{27} = 1$. The first thing to notice is which term is positive — here it’s the $y^2$ term. That tells us the transverse axis is vertical. In the standard form for a vertical hyperbola centred at the origin, we write:

where $a$ is the distance from the centre to each vertex (along the $y$-axis), and $b$ relates to the asymptotes and the shape of the hyperbola. The foci lie further out along the same axis, at a distance $c$ from the centre, where $c^2 = a^2 + b^2$.

Now let’s extract $a$ and $b$ from the given equation.

1. Identify $a^2$ and $b^2$ Comparing $\frac{y^2}{9} - \frac{x^2}{27} = 1$ with $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we get:

2. Find $c$ For a hyperbola, $c^2 = a^2 + b^2$ (note: it’s plus, not minus — a common mistake if you confuse it with an ellipse).

3. Vertices Since the hyperbola is vertical, the vertices lie on the $y$-axis at $(0, \pm a)$:

4. Foci The foci are further out on the same axis, at $(0, \pm c)$:

5. Eccentricity

An eccentricity greater than 1 is characteristic of a hyperbola; here it’s quite large, meaning the hyperbola is relatively “open”.

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