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.

| List-I | List-II |
| (P) m = | 1. 1/2 |
| (Q) Max area of | 2. 4 triangle EFG is |
| (R) y0 = | 3. 2 |
| ( *) y1 = | 4. 1 |

Codes:

| | P | Q | R | S |
| (A) | 4 | 1 | 2 | 3 |
| (B) | 3 | 4 | 1 | 2 |
| (C) | 1 | 3 | 2 | 4 |
| (D) | 1 | 3 | 4 | 2 |

Answer 1

Correct option (a)

Explanation:

Tangent at F, yt = x + 4t2

Area of the triangle EFG = A

So area is maximum.

Maximum Area = (6/4 – 8/8) = 1/2 sq unit.

G = (0, 4t) = (0, 2)

Thus, y1 = 2 F = (x0, y0) = (4t2, 8t) = (1, 4)

So, y0 = 4

Also, the point (4t2, 8t) = (1, 4) satisfies the line

y = mx + 3.

So, 4 = m + 3

⇒ m = 1