Question

A line $L:y=mx+3$ meets y –axis at $E(0,3)$ and the arc of the parabola $y^2=16x,0\le y\le 6$ 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 lists:

List - I	List - II
P. m =	1. $\frac{1}{2}$
Q. Maximum area of $\Delta EFG$ is	2. 4
R. $y_0=$	3. 2
S. $y_1=$	4. 1

• A. P Q R S
  4 1 2 3
• B. P Q R S
  1 3 2 4
• C. P Q R S
  3 4 1 2
• D. P Q R S
  1 3 4 2

Answer 1

The correct option is A P Q R S

4 1 2 3

Parabola is $y^2=16x$
and line is $y=mx+3$
Solving,
$(mx+3{)}^2=16x$
$\Rightarrow m^2{x}^2+(6m–16)x+9=0$ ...(i)
Also, tangent at F is
$y{y}_0=8(x+x_0)$
So, $y_1=\frac{8x_0}{y_0}$ ...(ii)
Now, area of $\Delta EFG,$
$\Delta =\frac{1}{2}(3-y_1)\cdot x_0$
$=\frac{1}{2}(3-\frac{8x_0}{y_0})x_0$
$=\frac{1}{2}(3x_0-\frac{8x_0^2}{y_0})$
$=\frac{1}{2}(3x_0-\frac{8x_0^2}{4\sqrt{x_0}})$
For $\Delta$ is to be maximum,
$\frac{d\Delta }{d{x}_0}=0$
$\Rightarrow 3-2\times \frac{3}{2}\sqrt{x_0}=0$
$\Rightarrow x_0=1$
So, $y_0=4$
$y_1=\frac{8x_0}{y_0}=2$
From $y_0=m{x}_0+3\Rightarrow m=1$