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