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1)

A line L:y=mx+3 meets y-axis at E (0,3) and the arc of the parabola y2  =16 x ,  0y6 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 EFG  has a local maximum

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

 1352021378_m12.JPG


A) P:4,Q:1,R:2,S:3

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

Answer:

Option A

Explanation:

Here,   y2=16x,0y6

   1452021564_m2.JPG

Tangent at F,

                   yt=x+at2

 at     x= 0, y= at=4t

 Also , (4t2,8t ) satisfy

   y=mx+c

     8t=4mt2+3

    4mt28t+3=0

     Area of   = 12[03104t14t28t1]=124t2(34t)

A= 2[3t2-4t3]

        dAdt=2[6t12t2]=12t(2t1)

1452021344_m3.JPG

  

Maximum at t=1/2

 and   4mt28t+3=0

     m-4+3=0

    m=1

 G(0,4t)  G(0,2)

y1=2

 (x0,y0)  =(4 t2,8t)=(1,4)

 y0=4

 Area   =2(3412)=12