Find the equation of hyperbola with foci (+4,0) and length of the latus rectum is 12.
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Answers
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(+c,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= 6×2=12
put a^2 and b^ 2 value in equation
x^2/ 4- y^ 2/ 12= 1
3x^2- y^ 2= 12
Answer:
Step-by-step explanation:
et equation of hyperbola be
x^2/a^2-y^2-b^2=1
F(+c,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= 6×2=12
put a^2 and b^ 2 value in equation
x^2/ 4- y^ 2/ 12= 1
3x^2- y^ 2= 12
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