Question

15.Foci(0,tVT0)passingthrough(2.3)

Answer 1

The given coordinates of focus are ( 0,± 10 ) and hyperbola passes through ( 2,3 ) .

Since the foci are on the y axis, the equation of the hyperbola is represented as,

y 2 a 2 − x 2 b 2 =1 ,where y is the transverse axis.(1)

Since y axis is the transverse axis, coordinates of Foci = (0,±c)

∴c= 10

a 2 + b 2 = c 2 a 2 + b 2 = 10 2 b 2 =100− a 2 (2)

Also, ( 2,3 ) satisfies the equation (1).

9 a 2 − 4 b 2 =1 9 a 2 − 4 10− a 2 =1 9( 10− a 2 )−4 a 2 = a 2 ( 10− a 2 ) .

Further simplifying the equations,

90−9 a 2 −4 a 2 =10 a 2 − a 4 a 4 −23 a 2 +90=0 a 4 −18 a 2 −5 a 2 +90=0 a 2 ( a 2 −18 )−5( a 2 −18 )=0 ( a 2 −18 )( a 2 −5 )=0

Hence, a 2 =5 or a 2 =18

In hyperbola, c>a i.e. c 2 > a 2 .Therefore a 2 =5

Substitute the value of a 2 in equation (2)

b 2 =10− a 2 b 2 =5

Now, substitute the values of a and b in equation (1) , we get

y 2 5 − x 2 5 =1

Thus, equation of hyperbola passing through ( 2,3 ) and foci ( 0,± 10 ) is y 2 5 − x 2 5 =1 .