Q.Find the equation of the hyperbola satisfying the given conditions: Vertices $(0, \pm 5)$, foci $(0, \pm 8)$.

The key idea is that the vertices and foci lie on the $y$-axis, so the hyperbola has a vertical transverse axis. Its standard form is

Step 1: From the vertices $(0, \pm 5)$, we have $a = 5$, so $a^2 = 25$.

Step 2: From the foci $(0, \pm 8)$, we have $c = 8$, so $c^2 = 64$. …

The first thing to notice is the coordinates of the vertices and foci. Both are given as $(0, \pm 5)$ and $(0, \pm 8)$. The $x$-coordinate is zero in every case. That tells you the centre of the hyperbola is at the origin $(0,0)$, and the transverse axis — the line that goes through the two vertices and the two foci — is the $y$-axis.

When the transverse axis is vertical, the standard equation of a hyperbola centred at the origin is:

Here, $a$ is the distance from the centre to each vertex, and $c$ is the distance from the centre to each focus. The relationship between $a$, $b$, and $c$ for a hyperbola is $c^2 = a^2 + b^2$. This is different from an ellipse, where $c^2 = a^2 - b^2$ — a common mix-up.

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

   The vertices are $(0, \pm 5)$, so $a = 5$.

   The foci are $(0, \pm 8)$, so $c = 8$.

2. Find $b^2$ using the hyperbola relation.

3. Write the equation. Since the $y$-term comes first (vertical axis), substitute $a^2 = 25$ and $b^2 = 39$: …