If `ea n de '` the eccentricities of a hyperbola and its conjugate, prove that `1/(e^2)+1/(e '^2)=1.`

Options

A. 0
B. 1
C. 2
D. 3

Answer

B

Explanation

Let the equation of hyperbola and conjugate hyperbola be
` ( x^ 2 ) /( a ^ 2 ) - ( y ^ 2 ) /( b ^ 2 ) = 1 and ( x ^ 2 ) /( a ^ 2 ) - ( y^ 2 ) / ( b ^ 2 ) = - 1 `
Then, the eccentricities are
` e ^ 2 = ( a ^ 2 + b ^ 2 ) /( a ^ 2 ) and e'^ 2 = ( a ^ 2 + b ^ 2 ) /( b ^ 2 ) `
` therefore ( 1 ) /( e^2 ) + ( 1 ) /( e'^ 2 ) = ( a ^ 2 ) /( a ^ 2 + b ^ 2 ) + ( b ^ 2 )/( a ^ 2 + b ^ 2 ) = 1 `