AP EAMCET Engineering 2013 Solved Paper

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)=

−3

2√2

(x−9)
⇒2√2y+24=−3x+27
⇒3x+2√2y−3=0