Question

find the equation of the hyperbola with foci (±4,0) and length of the latus rectum is 12​

Answer 1

Answer:

Answer:3x^2-y^2=12

Step-by-step explanation:

let equation of hyperbola be

x^2/a^2-y^2-b^2=1

F(+0,0) = F(+4,0)

Therefore, c=4 and c^2= 16

12=lr(length)=2b^2/a= b^2= 6a

c^2= a^2+ b^2

16-a^2+ 6a

a^2+ 6a-16=0

Therefore, a=- 8, 2 but a cannot be negative.

a=2 Therefore, a^2= 4

and b^2=6a= 6x2=12

put a^2 and b^2 value in equation

x^2/ 4- y^2/ 12= 1

3x^2- y^2= 12