If 3 , are the eccentric angles of the ends of a focal chord of the ellipse x^2 16 + y^2 12 =1 , then =

Step-by-step solution

Given ellipse, x^2 16 + y^2 12 =1 Let eccentricity of ellipse be ' e ' Then b^2=a^2 (1-e^2 ) Here, b^2=12, a^2=16 12=16 (1-e^2 )1-e^2= 3 4 or e^2=1- 3 4 = 1 4 or e= 1 2 If , are the eccentric angles of the ends of a focal chord of the ellipse, then eccentricity is given by e= ( - 2 ) ( + 2 ) Here, = 3 , = , and e= 1 2 1 2 = ( / 3- 2 ) ( / 3+ 2 ) Multiplying in N^r and D^r by 2 ( / 3+ 2 ) 1 2 = 2 ( / 3+ 2 ) ( / 3- 2 ) 2 ( / 3+ 2 ) ( / 3+ 2 ) = ( / 3+ + / 3- 2 )+ ( / 3+ - / 3+ 2 ) ( / 3+ ) 2 A B= ( A+B 2 )+ ( A-B 2 )