Given that, x = 9 is a chord of contact of hyperbola. x2−y2=9 ......(i) put x = 9, 81−y2=9 ⇒ y2=72 ⇒ y=6√2 or −6√2 ∴ Points are (9,6√2) and (9,−6√2) Now, differentiating Eq. (i) w.r.t. x, we get 2x−2y
dy
dx
=0 ⇒
dy
dx
=
x
y
at (9,6√2)(
dy
dx
)(9,6√2)=
9
6√2
=
3
2√2
and at (9−6√2)(
dy
dx
)(9,−6√2)=−
3
2√2
∴ Equation of tangent at (9,6√2) is (y−6√2)=
3
2√2
(x−9) ⇒2√2y−24=3x−27 ⇒3x−2√2y−3=0 and equation of tangent at (9,−6√2) is (y+6√2)=