Q.Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $5y^2 - 9x^2 = 36$.

Concept: Hyperbola with a vertical transverse axis, $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$.

Step 1 — Standard form. Divide $5y^2 - 9x^2 = 36$ by $36$:

so $a = \dfrac{6}{\sqrt5}$, $b = 2$ (positive $y^2$ term ⇒ vertical axis).

Step 2 — Find $c$. $c^2 = a^2 + b^2 = \dfrac{36}{5} + 4 = \dfrac{56}{5}$, so $c = \dfrac{2\sqrt{14}}{\sqrt5}$.

Step 3 — Read off the quantities.

• Vertices: $\left(0, \pm \dfrac{6}{\sqrt5}\right)$
• Foci: $\left(0, \pm \dfrac{2\sqrt{14}}{\sqrt5}\right)$ …

We are given the hyperbola $5y^2 - 9x^2 = 36$ and must find its foci, vertices, eccentricity, and latus-rectum length. The plan is to bring the equation to standard form, read off $a^2$ and $b^2$, then use the standard hyperbola relations.

Step 1 — Convert to standard form

Divide both sides by $36$ so the right side becomes $1$:

The positive term is the $y^2$ term, so the transverse axis is vertical (along the y-axis). Comparing with $\dfrac{y^2}{a^2} - \dfrac{x^2}{b^2} = 1$:

so $a = \dfrac{6}{\sqrt5}$ and $b = 2$.

Step 2 — Find $c$

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

So

Step 3 — Vertices and foci

Step 4 — Eccentricity

The $\sqrt5$ cancels, leaving $e = \dfrac{\sqrt{14}}{3}$ (which is $>1$, as it must be for a hyperbola). …