Question

find the equation of the hyperbola satisfying the given condition:

 foci (± \[3\sqrt{5}\]  0), the latus-rectum = 8

Answer 1

The foci of the hyperbola are \[\left( \pm 3\sqrt{5}, 0 \right)\] and the latus rectum is 8.
Thus, the value of  \[ae = 3\sqrt{5}\]

and \[\frac{2 b^2}{a} = 8\]

\[ \Rightarrow b^2 = 4a\]

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

\[\Rightarrow 4a = 45 - a^2 \]

\[ \Rightarrow a^2 + 4a - 45 = 0\]

\[ \Rightarrow \left( a - 5 \right)\left( a + 9 \right) = 0\]

\[ \Rightarrow a = - 9, 5\]

\[b^2 = - 36 \text { or }20\]

Since negative value is not possible, it is equal to 20.
Thus, the equation of the hyperbola is \[\frac{x^2}{25} - \frac{y^2}{20} = 1\].