Question

43.14.vertices (±7,0), e=

Answer 1

The given coordinates of vertices are ( ±7,0 ) and eccentricity is 4 3 .

Since the vertices are on the x axis, the equation of the hyperbola is represented as,

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

Since x axis is the transverse axis, coordinates of vertices = (±a,0)

∴a=7

Eccentricity = c a

So, c a = 4 3 c 7 = 4 3 c= 28 3

a 2 + b 2 = c 2 7 2 + b 2 = ( 28 3 ) 2 b 2 = 784 9 −49 b 2 = 784 9 −49 b 2 = 784−441 9 = 343 9

Substitute the values of a and b in equation (1)

x 2 49 − 9 y 2 343 =1

Thus, the equation of hyperbola with vertices ( ±7,0 ) and e= 4 3 is x 2 49 − 9 y 2 343 =1 .