The hyperbola can be written as:x²/ (144/25) + y²/ (81/25) = 1Now taking out e by using b²=a²( e² - 1 ) where a²= 144/25 and b²= 81/25....e comes out to be 5/4......focii of hyperbola are ( -+ 3, 0 )....this should also be the focii of ellipse....Now there are 2 cases when b² is less than 16 and other when b² is greater than 16.....Case 1:::::: 4e` = 3....gives e` = 3/4...b² = 16( 1- e`² ) which gives b² = 7.........Case 2:::::: be` = 3....gives e` = 3/b...16 = b²( 1 - e`² ) which gives b² = 25........Hence two values of b²...7 & 25
