Q.Find the equation of the hyperbola where foci are $(0, \pm 12)$ and the length of the latus rectum is $36$.

The key idea is that the foci lie on the $y$-axis, so the hyperbola is vertical with centre at the origin. Its standard form is

Step 1: For a vertical hyperbola, $c = 12$ (distance from centre to each focus). The relationship is $c^{2} = a^{2} + b^{2}$, so

Step 2: The length of the latus rectum for a vertical hyperbola is $\frac{2b^{2}}{a} = 36$. Hence

$b^{2} = 18a.$ …

The first thing to notice is where the foci lie. They are at $(0, \pm 12)$, which means the transverse axis is vertical — the hyperbola opens upward and downward. This immediately tells us the standard form we need.

For a hyperbola with a vertical transverse axis, the standard equation is:

Here $a$ is the distance from the centre to each vertex (along the $y$-axis), and $b$ relates to the conjugate axis. The foci are at $(0, \pm c)$, where $c$ is given by $c^2 = a^2 + b^2$. From the problem, $c = 12$, so:

Now the latus rectum. For any hyperbola, the length of the latus rectum (the chord through a focus, perpendicular to the transverse axis) is $\frac{2b^2}{a}$. The problem states this length is $36$, so:

We now have two equations in $a$ and $b^2$. Substitute (2) into (1):

This is a quadratic in $a$:

Solve it:

Since $a > 0$, we take the positive root:

Then from (2): …