If e and e^ are the eccentricities of the ellipse 5 x^2+9 y^2=45 and the hyperbola 5 x^2-4 y^2=45 respectively, then e e^ is equal to

Step-by-step solution

Equation of ellipse is or array r 5 x^2+9 y^2=45 x^2 9 + y^2 5 =1 e= 1-b^2 a^2 = 1-5 9 = 2 3 array and equation of hyperbola is 5 x^2-4 y^2=45 or x^2 9 - y^2 ( 45 4 ) =1 e^ = 1+ b^2 a^2 = 1+ 45 36 = 81 36 = 9 6 = 3 2 Then, e e^ = 2 3 3 2 =1