A line L: $y = m x + 3$ meets y-axis at E(0,3) and the arc of the parabola $y^{2} = 16 x, 0 \leq y \leq 6$ at the point $F (x_{0}, y_{0})$ . The tangent to the parabola at $F (x_{0}, y_{0})$ intersect the y-axis at $G (0, y_{1})$ . The slope m of the line L is chosen such that the area of the triangle EFG has local maximum.
Match ListI with List II and select the correct answer using the code given below lists:

List-I
List-II
(P) m= (1) $\frac{1}{2}$
(Q) Maximum area of $Δ E F G$ is (2) 4
(R) $y_{0} =$ (3) 2
(S) $y_{1} =$ (4) 1
(5) $\frac{1}{4}$

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P $\to$ 4; Q $\to$ 1; R $\to$ 2; S $\to$ 3

P $\to$ 3; Q $\to$ 4; R $\to$ 5; S $\to$ 2

P $\to$ 1; Q $\to$ 3; R $\to$ 2; S $\to$ 5

P $\to$ 1; Q $\to$ 3; R $\to$ 4; S $\to$ 2

answer is A.

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Detailed Solution

$Tangent at F, y t = x + 4 t^{2}$ , $Area of the triangle E F G = A$

$\begin{aligned} = \frac{1}{2} |\begin{array}{ccc} 0 & 3 & 1 \\ 0 & 4 t & 1 \\ 4 t^{2} & 8 t & 1 \end{array}| \\ = \frac{1}{2} 4 t^{2} (3 - 4 t) \\ = 2 t^{2} (3 - 4 t) \\ \Rightarrow A = (6 t^{2} - 8 t^{3}) \\ \Rightarrow \frac{d A}{d t} = 12 t - 23 t^{2} \\ \Rightarrow \frac{d^{2} A}{d t^{2}} = 12 - 48 t \end{aligned}$