Latika Leekha
Last Activity: 11 Years ago
Let the equation of hyperbola is x2/a2 – y2/b2 = 1.
where b2 = a2 (e2 -1).
For hyperbola, the eccentricity e > 1.
The focus is given as (4, 0). Hence, we have ae = 4.
This gives a = 4/e. … (1)
Latus Rectum = 2b2/ a = 12.
Hence, b2 = 6a. …... (2)
since, b2 = a2 (e2 -1)
Using (1) and (2) in the above equation, we get
6a = a2(e2 -1)
4/e = (4/e)2 (e2 -1)
Simplifying, we get
4e2 – 6e – 4 = 0
(2e +1)(e – 2) = 0.
This gives e = -1/2 or 2.
Hence, we have e = 2. ( since e > 1 in case of hyperbola).
hence, we have a = 4/e = 4/2 = 2.
Substituting this value we get b2 = 6a = 6.2 = 12
Hence, the eqauation of hyperbola is x2/a2 – y2/b2 = 1
x2/4 – y2/12 = 1.
Simplifying we get, the required hyperbola as
3x2– y2 = 12.
Thanks & Regards
Latika Leekha
askIITians Faculty
