Let F be (at2,2at)
F→(4t2,8t)
E→(0,3)
G→(0,y1)
y2=16x
Differentiating both the sides, we get
2yy′=16
⇒y′=8y where y=8t
⇒y′=1t
Now, the equation of the line passing through G is
⇒y=xt+c
c=am=4t ...(Tangent condition of parabola)
⇒y=xt+4t
∴y0=4t
G→(0,4t)
Area of triangle = 12∣∣
∣∣x1y11x2y21x3y31∣∣
∣∣=12∣∣
∣∣03104t14t28t1∣∣
∣∣
⇒A(t)=6t2−8t3
A′(t)−12t−24t2=0
⇒t=12
A(t)=6(12)2−8(12)3=32−1=12
G comes out to be (0,2)⇒y1=2
So, F(1,4)
Put F in y=mx+3 since it passes through it.
∴m=1