Let L ( x ₁, 4 ) be the end of the Latus rectum of the hyperbola x^2 a^2 - y^2 b^2 =1 lying in the first quadrant and let S (8, y₁ ) be the focus of the given hyperbola. Then the length of its transverse axis is

Step-by-step solution

We know that end of the latus rectum in lying in the first quadrant be (ae, .b^2 / a ) and focus be (c, 0) of hyperbola x ^2 a ^2 - y ^2 ~b ^2 =1 According to question b^2 a =4 b^2=4 a and c =8 We have c^2=a^2+b^2 64=a^2+49 a^2+4 a-64=0 aligned & a = -4 272 2 = -4 4 17 2 & a = 4 17 -4 2 [Take + sign because end of latus rectium & lies in first quadrant] & Length of transverse axis =2 a =4( 17 -1) aligned