VITEEE 2015 Paper

ϕm (m) = 3 + 2m - 7m2+2m3
Φ2 (m) = -14 m + 7n2
Φ3′ (m) = 2 - 14m + 6m2
Now, putting ϕm (m) = 0 , we have
3 + 2m - 7m2+2m3 = 0
⇒ (1 - m) (1 + 2m) (3 - m) = 0
⇒ m = -

1

2

, 1 , 3
We know that
cϕn′ (m) + Φn−1 (m) = 0 , which in the given case becomes
c (2 - 14m + 6m2) + (- 1 4m + 7m2) = 0
⇒ c =

14m−7n2

2−14m+6m2

So, when m = -

1

2

, c = −

5

6

When m = 1, c = −

7

6

When m = 3, c = −

3

2

∴ Asymptotes are y = −

1

2

x−

5

6

, y = x -

7

6

and y = 3x -

3

2