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Question

A line L:y=mx+3 meets y-axis at E(0,3) and the arc of the parabola y2=16x,0y6 at the point F(x0,y0). The tangent to the parabola at F(x0,y0) intersects the y-axis at G(0,y1). The slope m of the line L is chosen such that the area of the EFG has a local maximum.
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Solution

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)=6t28t3
A(t)12t24t2=0
t=12
A(t)=6(12)28(12)3=321=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

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