Question

find the equation of the hyperbola satisfying the given condition:

foci (0, ± 12), latus-rectum = 36

Answer 1

The foci of the hyperbola are \[\left( 0, \pm 12 \right)\] and the latus rectum is 36.
Thus, the value of  \[ae = 12\].

and \[\frac{2 b^2}{a} = 36\]

\[ \Rightarrow b^2 = 18a\]

Now, using the relation 

\[b^2 = a^2 ( e^2 - 1)\],we get:

\[\Rightarrow 18a = 144 - a^2 \]

\[ \Rightarrow a^2 + 18a - 144 = 0\]

\[ \Rightarrow \left( a + 24 \right)\left( a - 6 \right) = 0\]

\[ \Rightarrow a = - 24, \text { or }6\]

\[b^2 = - 432 \text { or }108 \](but negative value is not possible)\]

Thus, the equation of the hyperbola is

\[\frac{y^2}{36} - \frac{x^2}{108} = 1\].