ParabolaHard
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. 1/2
Q. Maximum area of ᐃEFG is 2. 4
R. y0 = 3. 2
S. y1 = 4. 1
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. 1/2
Q. Maximum area of ᐃEFG is 2. 4
R. y0 = 3. 2
S. y1 = 4. 1
Options
A.P → 4, Q → 1, R → 3, S → 2
B.P → 3, Q → 4, R → 1, S → 2
C.P → 1, Q → 3, R → 2, S → 4
D.P → 1, Q → 3, R → 4, S → 2
Solution
Let F(4t2, 8t)
where
ᐃ = (6t2 - 8t3)
y0 = 8t = 4 & y1 = 4t = 2
where
ᐃ = (6t2 - 8t3)
y0 = 8t = 4 & y1 = 4t = 2
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