Q.Find the eccentricity of the hyperbola $9y^2 - 4x^2 = 36$.

The key idea is that the eccentricity $e$ of a hyperbola depends on its standard form.

First, rewrite the given equation in standard form:

This is a vertical hyperbola (the $y^2$ term is positive) with $a^2 = 4$ and $b^2 = 9$. …

The first thing to notice is that the $y^2$ term comes first and is positive. That tells us the hyperbola opens up and down — it's a vertical hyperbola. Many students rush to assume the standard form $x^2/a^2 - y^2/b^2 = 1$, but here the roles of $x$ and $y$ are swapped.

For a vertical hyperbola, the standard form is:

where $a$ is the distance from the centre to each vertex (along the vertical axis), and $b$ relates to the asymptotes. The eccentricity $e$ is always greater than 1, and for a vertical hyperbola it's given by:

Let's work through the problem step by step.

1. Rewrite the equation in standard form. We start with $9y^2 - 4x^2 = 36$. Divide every term by 36 to get 1 on the right:

Simplify each fraction:

2. Identify $a^2$ and $b^2$.

   In the form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, we match:

   • $a^2 = 4$ (under $y^2$), so $a = 2$.
   • $b^2 = 9$ (under $x^2$), so $b = 3$.
   Watch out

   A common mistake is to take $a^2 = 9$ because 9 is larger. But $a$ always belongs to the positive term. Here $y^2$ is positive, so $a^2$ is the denominator under $y^2$, which is 4.

3. Apply the eccentricity formula.

   For any hyperbola, $e = \sqrt{1 + \frac{b^2}{a^2}}$. Substitute $a^2 = 4$ and $b^2 = 9$: …