If `alpha and beta` are eccentric angles of the ends of a focal chord of the ellipse `x^2/a^2 + y^2/b^2 =1`, then `tan alpha/2 .tan beta/2` is (A) `(1-e)/(1+e)` (B) `(e+1)/(e-1)` (C) `(e-1)/(e+1)` (D) none of these

IF alpha , beta are eccentric angles of end points of a focal chord of the ellipse x^(2)/a^(2) + y^(2)/b^(2) =1 then tan(alpha /2) tan (beta/2) is equal to

If alpha,beta are the eccentric angles of the extremities of a focal chord of the ellipse x^(2)/16 + y^(2)/9 = 1 , then tan (alpha/2) tan(beta/2) =

If alpha,beta are the ends of a focal chord of the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 then its eccentricity e is

IF alpha,beta are the eccentric angles of the extremities of a focal chord of the ellipse x^2/a^2+y^2/b^2=1 . Then show that e cos""(alpha+beta)/2=cos""(alpha-beta)/2

If alpha and beta be the eccentric angles of the extremities of a focal chord of the hyperbola b^(2)x^(2) - a^(2)y^(2) = a^(2)b^(2) , show that, tan(alpha)/(2)tan(beta)/(2) = -(e-1)/(e+1) , (e-1)/(e+1) (e is the eccentricity of the hyperbola.).

If α, β are the eccentric angles of the extremeties of a focal chord of the ellipse (i) e cos (alpha+beta)/(2)=cos (alpha-beta)/(2) (ii) tan (alpha/2)tan (beta/2)=(e-1)/(e+1)

If the eccentric angles of the extremities of a focal chord of an ellipse x^2/a^2 + y^2/b^2 = 1 are alpha and beta , then (A) e = (cos alpha + cos beta)/(cos (alpha + beta)) (B) e= (sin alpha + sin beta)/(sin(alpha + beta)) (C) cos((alpha-beta)/(2)) = e cos ((alpha + beta)/(2)) (D) tan alpha/2.tan beta/2 = (e-1)/(e+1)

If (asectheta;btantheta) and (asecphi; btanphi) are the ends of the focal chord of x^2/a^2-y^2/b^2=1 then prove that tan(theta/2)tan(phi/2)=(1-e)/(1+e)

If (asectheta;btantheta) and (asecphi; btanphi) are the ends of the focal chord of x^2/a^2-y^2/b^2=1 then prove that tan(x/a)tan(phi/2)=(1-e)/(1+e)