Given: Foci (0, ±12), the latus-rectum = 36
To find: equation of the hyperbola
Formula used:
The standard form of the equation of the hyperbola is,
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = -1\)
Coordinates of the foci for a standard hyperbola is given by (0, ±be)
Length of latus rectum is \(\frac{2a^2}{b}\)
According to the question:
![](https://www.sarthaks.com/?qa=blob&qa_blobid=14781495449133173736)
We know,
a2 = b2(e2 – 1)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5538394519311881422)
⇒ 18b = 144 – b2
⇒ b2 + 18b – 144 = 0
⇒ b2 + 24b – 6b – 144 = 0
⇒ b(b + 24) – 6(b + 24) = 0
⇒ (b + 24)(b – 6) = 0
⇒ b = -24 or b = 6
Since b is a distance, and it can’t be negative
⇒ b = 6
⇒ b2 = 36 a2 = 18b
⇒ a2 = 18(6)
⇒ b2 = 108
Hence, equation of hyperbola is:
![](https://www.sarthaks.com/?qa=blob&qa_blobid=5581462001106968796)