=1 Centre : O(0,0) Foci : S(±ae,0)≡(±3,0) ⇒‌‌ae=3 Eccentricity, e=√1+(‌
b
a
)2=‌
3
2
∵‌ae‌=3 ⇒a.(‌
3
2
)‌=3 ‌⇒‌‌a=2 and √1+(‌
b
a
)2=‌
3
2
‌⇒‌‌1+(‌
b
2
)2=‌
9
4
‌⇒‌‌‌
b2
4
=‌
5
4
‌⇒‌‌b2=5 ∴ Hyperbola equation is ‌
x2
4
−‌
y2
5
=1 .....(i) Given, line is 2x−y=1 Put y=2x−1 into Eq. (i), ‌‌
x2
4
−‌
(2x−1)2
5
=1 ‌⇒‌‌5x2−4(4x2+1−4x)=20 ‌⇒‌‌5x2−16x2−4+16x−20=0 ‌⇒‌‌−11x2+16x−24=0 Here, D=b2−4ac=(16)2−4(−11)(−24)<0 ∴ Line does not intersect the hyperbola.