If the angle between the asymptotes of the hyperbola x^2-k y^2=3 is 3 and e is its eccentricity, then the pole of the line x+y-1=0 with respect to this hyperbola is

Step-by-step solution

Given the equation of hyperbola is x^2-k y^2=3 x^2 3 - y^2 3 k =1 So, angle between the asymptotes is = ⁻¹ ( 2 a b a^2-b^2 ) ( 3 )= 2 3 3 k 3- 3 k aligned & 3 = 2 k k-1 3= 4 k (k-1)^2 k=3, 1 3 & So, e= a^2+b^2 a = 3+1 3 = 2 3 aligned Now, the hyperbola is x^2 3 - y^2 1 =1 So, pole of the line x+y-1=0 with respect to the hyperbola is ( -a^2 l n , b^2 m n ) aligned & = ( -3 1 -1 , 1 1 -1 )=(3,-1) & = (3,- 3 2 2 3 )= (k,- 3 2 e ) aligned