The equation of the hyperbola is x2−ky2=3 which can be rewritten in standard form as: ‌
x2
3
−‌
y2
3
k
=1 Here, a2=3 and b2=‌
3
k
. The angle between the asymptotes is given by tan−1‌
b
a
=‌
Ï€
3
This implies: tan‌
Ï€
6
=‌
b
a
=‌
√
3
k
√3
=‌
1
√k
Therefore: ‌
1
√3
=‌
1
√k
⇒k=3 Recalculate the standard form with k=3 : 3x2−y2=9 The pole of the line x+y−1=0 with respect to the hyperbola 3x2−y2=9 can be found. Let (h,k) be t pole. The equation of the pole is defined using S1=0 : 3hx−ky=9 This equation is rewritten as: ‌
hx
3
−‌
ky
3
=1 Comparing with the equation x+y−1=0, we get: ‌
h
3
=1‌‌‌ and ‌‌‌‌
−k
3
=1 Solving these gives: h=3‌‌‌ and ‌‌‌k=−9 Thus, the pole of the line is (3,−9). The final expression representing the pole in terms of a parameter is: (k,‌