WBJEE 2016 Math Solved Paper

Let (

1

2

cosθ,

1

3

sinθ) be a point on 4x2 + 9y2 = 1, so equation of tangent at (

1

2

cosθ,

1

3

sinθ) is
2x cos θ + 3 y sin θ = 1
equating slope with 8x = 9y

−2cosθ

3sinθ

=

8

9

⇒ tan θ = -

3

4

Hence either cos θ = -

4

5

, sin θ =

3

5

or cos θ =

4

5

, sin θ = -

3

5

so the points are (−

2

5

,

1

5

) or (

2

5

,−

1

5

)