Given: Foci (±4, 0), the latus-rectum = 12
To find: equation of the hyperbola
Formula used:
Standard form of the equation of hyperbola is,
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\)
Coordinates of the foci for a standard hyperbola is given by (±ae, 0)
Length of latus rectum is \(\frac{2b^2}{a}\)
According to the question:
![](https://www.sarthaks.com/?qa=blob&qa_blobid=8221509982736745707)
We know,
b2 = a2(e2 – 1)
![](https://www.sarthaks.com/?qa=blob&qa_blobid=17085136991949545873)
⇒ 6a = 16 – a2
⇒ a2 + 6a – 16 = 0
⇒ a2 + 8a – 2a – 16 = 0
⇒ a(a + 8) – 2(a + 8) = 0
⇒ (a + 8)(a – 2) = 0
⇒ a = -8 or a = 2
Since a is a distance, and it can’t be negative,
⇒ a = 2
⇒ a2 = 4 b2 = 6a
⇒ b2 = 6(2)
⇒ b2 = 12
Hence, equation of hyperbola is:
![](https://www.sarthaks.com/?qa=blob&qa_blobid=12738425593882836116)