Given, conjugate lines are 2x+3y+12=0 . . . (i) and x−y+k=0 . . . (ii) We know, that two lines are said to be conjugate with respect to a curve, if each passes through the pole of the polar of that curve. Let (x1,y1) be the pole of parabola It's polar is ‌‌yy1=4(x+x1) ⇒‌4x−y1y+4x1=0 ⇒2x−(‌
y1
2
)y+2x1=0 . . . (iii) On comparing Eqs. (i) and (iii), we get ‌
−y1
2
=3⇒y1=−6 and 2x1=12⇒x1=6 ∴‌‌ Pole (x1,y1)=(6,−6) Eq. (ii) also passes through pole (6,−6) ∴‌6−(−6)+k=0 ⇒k=−12