Question

find the equation of hyperbola with the centre at the origin ,transverse axis 12 and one of the foci at (3,√5,0)​

Answer 1

Answer :

Substituting ( 2 ) in ( 1 ) we get,

Substituting ( 2 ) in ( 1 ) we get,⇒ a

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45⇒ a

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45⇒ a 2

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45⇒ a 2 +4a−45=0

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45⇒ a 2 +4a−45=0⇒ a

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45⇒ a 2 +4a−45=0⇒ a 2

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45⇒ a 2 +4a−45=0⇒ a 2 +9a−5a−45=0

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45⇒ a 2 +4a−45=0⇒ a 2 +9a−5a−45=0⇒ a(a+9)−5(a+9)=0

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45⇒ a 2 +4a−45=0⇒ a 2 +9a−5a−45=0⇒ a(a+9)−5(a+9)=0⇒ (a+9)(a−5)=0

Substituting ( 2 ) in ( 1 ) we get,⇒ a 2 +4a=45⇒ a 2 +4a−45=0⇒ a 2 +9a−5a−45=0⇒ a(a+9)−5(a+9)=0⇒ (a+9)(a−5)=0So, a=5 or a=−9