A line L : y = mx + 3 meets y-axis at E(0, 3) and the arc of the parabola y2 = 16x, 0 ≤ y ≤ 6 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 triangle EFG has a local maximum. Match List-I with List-II and select the correct answer using the code given below the lists: List I List II (P) m = (1) \frac{1}{2} (Q) Maximum area of ΔEFG is (2) 4 (R) y0 = (3) 2 (S) y1 = (4) 1

Engineering

Mathematics

Tangent of Parabola

Question

A line L : y = mx + 3 meets y-axis at E(0, 3) and the arc of the parabola y² = 16x, 0 ≤ y ≤ 6 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 triangle EFG has a local maximum.

Match List-I with List-II and select the correct answer using the code given below the lists:

List I	List II
(P) m =	(1) $\frac{1}{2}$
(Q) Maximum area of ΔEFG is	(2) 4
(R) y₀ =	(3) 2
(S) y₁ =	(4) 1

P → 4

Q → 1

R → 2

S → 3

P → 3

Q → 4

R → 1

S → 2

P → 1

Q → 3

R → 4

S → 2

P → 1

Q → 3

R → 2

S → 4

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Now, $∆$ = area of ( $∆$ EFG)

$\frac{1}{2} | | \begin{matrix} 0 & 3 & 1 \\ 0 & 4 t & 1 \\ 4 t^{2} & 8 t & 1 \end{matrix} | |$

$∆$ = | 6t² – 8t³ |

$∴ \frac{dΔ}{dt} = 0 \Rightarrow t = 0, \frac{1}{2}$ , but t = 0 (reject)

So, F (x₀, y₀) $\equiv$ (4t², 8t) $\equiv$ (1, 4)

E (0, 3)

G (0, y₁) $\equiv$ (0, 4t) $\equiv$ (0, 2)

Now, y = mx + 3, satisfied by (1, 4) ⇒ 4 = m + 3 ⇒ m = 1.