JEE Mains Model Paper 4

Given equation of lines are:
√3x - y - 4√3k = 0 (i)
√3kx + ky - 4√3 = 0 (ii)
From eq 1 k =

√3x−y

4√3

From eq 2 k =

4√3

√3x+y

Equating the values of k, we get

√3x−y

4√3

=

4√3

√3x+y

⇒ 3x2−y2 = 48 ⇒

x2

16

−

y2

48

= 1
which is equation of hyperbola
∴ a2 = 1 and b2 = 48
⇒ e2 = 1 +

48

16

= 1 + 3 = 4
⇒ e = 2