Q.Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 4, 0)$, the latus rectum is of length $12$.

Concept: Standard form of a hyperbola with foci on the $x$-axis.

For a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with foci on the $x$-axis, the foci are at $(\pm c, 0)$ where $c^2 = a^2 + b^2$, and the length of the latus rectum is $\frac{2b^2}{a}$.

Step 1: From foci $(\pm 4, 0)$, we have $c = 4$, so $c^2 = 16$.

Step 2: From latus rectum length $12$, we get $\frac{2b^2}{a} = 12$, which gives $b^2 = 6a$.

Step 3: Using $c^2 = a^2 + b^2$:

$16 = a^2 + 6a$ …

The foci lie on the $x$-axis at $(\pm 4, 0)$, which tells us immediately that this is a horizontal hyperbola centered at the origin. The standard form for such a hyperbola is

where the foci are located at $(\pm c, 0)$ with $c^2 = a^2 + b^2$. The key is to extract both $a$ and $b$ from the two pieces of information given.

Finding the parameters

1. Extract $c$ from the foci.

Since the foci are at $(\pm 4, 0)$, we have $c = 4$. This gives us the fundamental relationship:

2. Use the latus rectum to get a second equation.

The latus rectum of a hyperbola is the chord through a focus perpendicular to the transverse axis. For a horizontal hyperbola, its length is given by:

We're told this length is $12$, so:

Simplifying:

3. Solve the system of equations.

We now have two equations in two unknowns:

• $a^2 + b^2 = 16$
• $b^2 = 6a$

Substitute the second into the first:

Factor this quadratic:

$(a + 8)(a - 2) = 0$ …