IIT JEE Advanced 2012 Paper 1

Slope of tangent = 2.
The tangents are y = 2x±√9×4−4
That is, 2x - y = ±4√2
⇒

x

2√2

−

y

4√2

= 1 and −

x

2√2

+

y

4√2

= 1
Comparing it with

xx1

9

−

yy1

4

= 1, we get the point of contact as (

9

2√2

,

1

√2

) and (−

9

2√2

,−

1

√2

).
Alternatslope Solution: Equation of tangent at P(θ) is
(

secθ

3

)×−(

tanθ

2

)y = 1
⇒ Slope =

2secθ

3tanθ

= 2
⇒ sin θ =

1

3

⇒ points are (

9

2√2

,

1

√2

) and (−

9

2√2

,−

1

√2

)