A line L:y=mx+3 meets y-axis at E(0,3) and the arc of the parabola y 2 =16x,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 △EFG has a local maximum. Match List 1 with List 2 List 1 List 2 A. m= 1. 2 1 ​ B. Maximum area of ΔEFG is 2. 4 C. y 0 ​ = 3. 2 D. y 1 ​ = 4. 1

Answer

`atos;bto p;ctoq;d to r`

Explanation

`atos,bto p,ctoq,d to r`
Let point F on the parabola be `(4t^(2),8t)`.
Tangent at this point is `ty=x+4t^(2)`
It meets the y-axis at (0,4t).
Then the area of triangle
EFG is `A(t)=2t^(2)(3-4t)=6t-8t^(3)`
Differentiating w.r.t, we get
`A'(t)=12t-24t^(2)`
For A'=0, t=1/2, which is a maxima. So, point F is (1,4).
slope of EF = 1
`:." "m=1orA(t)|_(max.)=(1)/(2)` sq. units
`y_(0)=4`
`andy_(1)=2`