Let the foci and length of the latus rectum of an ellipse x2a2+ y2b2=1, a>b be ( 5, 0) and 50, respectively. Then, the square of the eccentricity of the hyperbola x2b2- y2a2b2=1 equals
Answer
51
Solution
a2-b2=(ae)2=25 and 2b2a= 50=5 2 ⇒ b2= 5a 2
a2- 5a 2-25=0 ⇒ (a-5 2 ) (a+ 5 2 )=0 ⇒ a=5 2, b2=25
Hyperbola: e12=1+ a2b2b2=1+a2=1+50=51
Answer: 51