Q.The equation of the hyperbola with vertices at $(0, \pm 6)$ and eccentricity $\dfrac{5}{3}$ is ________ and its foci are ________.

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

Step 1 — Identify $a$ and $c$

Vertices at $(0, \pm 6)$ give $a = 6$.

Eccentricity $e = \frac{5}{3}$, so $c = a e = 6 \cdot \frac{5}{3} = 10$.

Step 2 — Find $b^2$

For a hyperbola, $c^2 = a^2 + b^2$.

Thus $b^2 = c^2 - a^2 = 100 - 36 = 64$, so $b = 8$.

Step 3 — Write equation and foci …

Why this approach works

The vertices are given as $(0, \pm 6)$. That tells you two things immediately: the centre is at the origin, and the transverse axis is vertical (because the vertices lie on the y-axis). For a hyperbola centred at the origin with a vertical transverse axis, the standard equation is

where $a$ is the distance from the centre to each vertex. Here $a = 6$.

The eccentricity $e$ is defined as $e = \frac{c}{a}$, where $c$ is the distance from the centre to each focus. For any hyperbola, $c^2 = a^2 + b^2$. So once we know $a$ and $e$, we can find $c$, then $b$, and write the full equation.

Step-by-step solution

1. Identify $a$ from the vertices.

Vertices are $(0, \pm 6)$, so the distance from centre $(0,0)$ to either vertex is $6$. Hence

2. Use eccentricity to find $c$.

Eccentricity $e = \frac{5}{3}$ and $e = \frac{c}{a}$. Therefore

So the foci are at $(0, \pm 10)$.

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

For a hyperbola, $c^2 = a^2 + b^2$. Thus

So $b = 8$. …