Line L:3x−2y+12=0 Parabola P : 4y=3x2 By putting y=
3x2
4
in equation of line We get, 3x−2(
3x2
4
)+12=0 ⇒6x−3x2+24=0 ⇒x2−2x+8=0 ⇒x=4,−2 for x=4, we get y=12 for x=−2, we get y=3 So, points A and B are (4,12) and (−2,3)Now, Vertex of parabola is (0,0) ⇒tanθ=|