Given, lines are √3kx+ky−4√3=0. . . (i) √3x−y−4√3k=0‌‌...‌ (ii) ‌ Multiply Eq. (ii) ×k and then adding Eqs. (i) and (ii), √3kx+ky−4√3=0 ‌
√3kx−ky−4√3k2=0
(2√3x)k=4√3+4√3k2
∴‌‌x=‌
4√3(1+k2)
2√3k
=2(k+‌
1
k
) Subtracting Eq. (i) from Eq. (ii), √3kx+ky−4√3=0 √3kx−ky−4√3k2=0 ‌
−‌‌+‌‌+
2ky=4√3−4√3k2
y=‌
4√3(1+k2)
2k
=2√3(k+‌
1
k
) We have, x=2(k+‌
1
k
) and y=2√3(‌
1
k
−k) ‌
x
2
=(k+‌
1
k
). . . (iii) ‌
y
2√3
=(‌
1
k
−k) . . . (iv) Squaring and subtracting Eq. (iii) from Eq. (iv), ‌
x2
4
−‌
y2
12
‌‌=(k2+‌
1
k2
+2)−(‌
1
k2
+k2−2) ‌
x2
4
−‌
y2
12
‌‌=4 ‌ or ‌‌‌‌
x2
16
−‌
y2
48
‌‌=1 Clearly, this is a hyperbola. ∴‌‌e2‌‌=1+‌