JEE Advanced2013MathematicsApplication of DerivativesActual
A line L : y = m x + 3 meets y - a x is at E(0,3) and the arc of the parabola y 2 = 1 6 x , 0 ≤ y ≤ 6 at the point F ⁡ x 0 y 0 .The tangent to the parabola at F ⁡ x 0 y 0 intersects 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 a local maximum. Match List I with List II and select the correct answer using the code given below the
Options
- Aa-p;b-q;c-s;d-r;
- Ba-s;b-p;c-q;d-r;
- Ca-r;b-q;c-s;d-p;
- Da-p;b-q;c-s;d-r;
Correct answer
B. a-s;b-p;c-q;d-r;
Step-by-step solution
tangent at F yt = c + 4t 2 a : x = 0 y = 4t (0,4t) (4t 2 ,8t) satisfies the line 8t = 4mt 2 + 3 4mt 2 - 8t + 3 = 0 Area = 1 2 0 3 1 0 4 t 1 4 t 2 8 t 1 = 1 2 4 t 2 3 - 4 t = 2 t 2 3 - 4 t A = 2 3 t 2 - 4 t 3 = dA dt = 2 6 t - 1 2 t 2 = 24 t(1 - 2t) t = 1/2 maxima G 0 4 t ⇒ G 0 2 y 1 = 2 x 0 y 0 = 4 t 2 8 t = 1 4 y 0 = 4 Area = 2 3 4 - 1 2 = 2 3 - 2 4 = 1 2