Question

(vii)  find the equation of the hyperbola satisfying the given condition:

foci (± 4, 0), the latus-rectum = 12

Answer 1

 The foci of the hyperbola are \[\left( \pm 4, 0 \right)\] and the latus rectum is 12.
Thus, the value of  \[ae = 4\]

and \[\frac{2 b^2}{a} = 12\]

\[ \Rightarrow b^2 = 6a\]

Now, using the relation 

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

\[\Rightarrow 6a = 16 - a^2 \]

\[ \Rightarrow a^2 + 6a - 16 = 0\]

\[ \Rightarrow \left( a - 2 \right)\left( a + 8 \right) = 0\]

\[ \Rightarrow a = 2, or - 8\] 

\[b^2 = 12 \text { or }- 48\]

Since negative value is not possible, its value is 12.
Thus, the equation of the hyperbola is \[\frac{x^2}{4} - \frac{y^2}{12} = 1\].