Q.Find the equation of the hyperbola satisfying the given conditions: Foci $(0, \pm 13)$, the conjugate axis is of length $24$.

Concept: Hyperbola with vertical transverse axis (foci on the y-axis).

The standard form is

where $c^2 = a^2 + b^2$, foci at $(0, \pm c)$, and conjugate axis length $= 2b$.

Step 1: From foci $(0, \pm 13)$, we have $c = 13$.

Step 2: Conjugate axis length $= 24$, so $2b = 24 \Rightarrow b = 12$. …

The first thing to notice is where the foci lie. They are at $(0, \pm 13)$, which means they are on the $y$-axis. For a hyperbola, the foci always lie on the transverse axis — the axis that goes through the two branches. So here, the transverse axis is vertical (along the $y$-axis), and the conjugate axis is horizontal (along the $x$-axis).

That tells us the standard form we need. For a vertical transverse axis, the hyperbola's equation is:

Here $a$ is the distance from the center to each vertex (along the $y$-axis), and $b$ is the distance from the center to each endpoint of the conjugate axis (along the $x$-axis). The foci are at $(0, \pm c)$, where $c$ is related to $a$ and $b$ by $c^2 = a^2 + b^2$.

Now, the problem gives us two pieces of data: the foci and the length of the conjugate axis.

1. Find $c$ from the foci.

   The foci are $(0, \pm 13)$, so the distance from the center $(0,0)$ to each focus is $c = 13$.

2. Find $b$ from the conjugate axis length.

   The conjugate axis is the segment perpendicular to the transverse axis, passing through the center. Its length is $2b$. We are told this length is $24$, so:

3. Use the relationship $c^2 = a^2 + b^2$ to find $a^2$. For a hyperbola, $c$ is the largest of the three numbers. We have:

$a^2 = 25$ …