A line L : y = mx + 3 meets y-axis at E (0, 3) and the arc of the parabola `y^2 = 16x` `0leyle6` at the point art `F(x_0,y_0)`. The tangent to the parabola at `F(X_0,Y_0)` intersects the y-axis at `G(0,y)`. The slope m of the line L is chosen such that the area of the triangle EFG has a local maximum P) m= Q) = Maximum area of `triangle EFG` is (R) `y_0=` (S) `y_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`