If the latus rectum subtends a right angle at center of the hyperbola, then its eccentricity is

Step-by-step solution

Given, latus rectum of hyperbola subtends 90º at its centre. Let there be a hyperbola x^2 a^2 - y^2 b^2 =1 So, eccentricity, e= 1+ b^2 a^2 End points of latusrectum are L= (a e, b^2 a ) and L^ = (a e, -b^2 a ) Centre = (0, 0) L C L^ =90^ Then, in L C L^ , by using Pythagoras theorem, aligned (L C)^2+ (L^ C )^2 & = (L L^ )^2 (a e-0)^2 & + ( b^2 a -0 )^2+(a e-0)^2+ ( -b^2 a -0 )^2 & = ( 2 b^2 a )^2 aligned aligned & 2 (a^2 e^2+ b^4 a^2 )= 4 b^4 a^2 2 b^4 a^2 =2 a^2 e^2 & b^4=a^4 e^2 & [a^2 (e^2-1 ) ]^2=a^4 e^2 b^2=a^