Q.Find the equation of the hyperbola with foci $(0, \pm 3)$ and vertices $\left(0, \pm \frac{\sqrt{11}}{2}\right)$.

Concept: Hyperbola with vertical transverse axis (standard form $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$).

Step 1 – Identify orientation and parameters

Foci at $(0, \pm 3)$ and vertices at $\left(0, \pm \frac{\sqrt{11}}{2}\right)$ tell us the transverse axis is vertical.

Here $c = 3$ (distance from centre to each focus) and $a = \frac{\sqrt{11}}{2}$ (distance to each vertex).

Step 2 – Find $b^2$ using $c^2 = a^2 + b^2$

Step 3 – Write the equation …

The first thing to notice is where the foci and vertices lie. Both are on the y-axis: foci at $(0, \pm 3)$ and vertices at $(0, \pm \frac{\sqrt{11}}{2})$. That tells you the hyperbola opens upward and downward — its transverse axis is vertical.

For a vertical hyperbola centered at the origin, the standard equation is:

Here $a$ is the distance from the center to each vertex, and $c$ is the distance from the center to each focus. The relationship between $a$, $b$, and $c$ for a hyperbola is $c^2 = a^2 + b^2$ (note: for an ellipse it’s $c^2 = a^2 - b^2$, so don’t mix them up).

1. Identify $a$ and $c$ directly from the given points.

   The vertices are at $(0, \pm \frac{\sqrt{11}}{2})$, so $a = \frac{\sqrt{11}}{2}$.

   The foci are at $(0, \pm 3)$, so $c = 3$.

2. Use $c^2 = a^2 + b^2$ to find $b^2$.

3. Write the equation. Plug $a^2 = \frac{11}{4}$ and $b^2 = \frac{25}{4}$ into the standard form: …