Hyperbola

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

=‌

π

3

This implies:
tan‌

π

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,‌

−√3e

2

)
where k=3.