Q.Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 3\sqrt{5}, 0)$, the latus rectum is of length $8$.

Concept: Hyperbola with a horizontal transverse axis, $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$, foci $(\pm c, 0)$, latus rectum $\dfrac{2b^2}{a}$.

Step 1 — From the foci. $(\pm 3\sqrt5, 0) \Rightarrow c = 3\sqrt5 \Rightarrow c^2 = 45 = a^2 + b^2$.

Step 2 — From the latus rectum. $\dfrac{2b^2}{a} = 8 \Rightarrow b^2 = 4a$. …

We must find the hyperbola whose foci are $(\pm 3\sqrt5, 0)$ and whose latus rectum has length $8$. Both data connect directly to $a$, $b$, $c$ in the standard equation, so we set up the relations and solve.

Step 1 — Orientation and $c$

The foci $(\pm 3\sqrt5, 0)$ lie on the x-axis, symmetric about the origin, so the transverse axis is horizontal and the standard form is

From the foci, $c = 3\sqrt5$, so

Step 2 — Latus-rectum condition

For a horizontal hyperbola the latus-rectum length is $\dfrac{2b^2}{a}$. Given it equals $8$:

Step 3 — Solve the system

Substitute $b^2 = 4a$ from (2) into (1): …