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 $△EFG$ has a local maximum.
Match List 1 with List 2

	List 1		List 2
A.	$m=$	1.	$\frac{1}{2}$
B.	Maximum area of $\Delta EFG$ is	2.	$4$
C.	$y_0=$	3.	$2$
D.	$y_1=$	4.	$1$

• A. $A-4,B-2,C-1,D-3$
• B. $A-1,B-4,C-2,D-3$
• C. $A-4,B-1,C-2,D-3$
• D. $A-4,B-2,C-3,D-1$

Answer 1

The correct option is C $A-4,B-1,C-2,D-3$

Let $F$ be $(a{t}^2,2at)$
$F\to (4t^2,8t)$
$E\to (0,3)$
$G\to (0,y_1)$
$y^2=16x$
Differentiating both the sides, we get
$2y{y}^{\prime }=16$
$\Rightarrow y^{\prime }=\frac{8}{y}$ where $y=8t$
$\Rightarrow y^{\prime }=\frac{1}{t}$
Now, the equation of the line passing through G is
$\Rightarrow y=\frac{x}{t}+c$
$c=\frac{a}{m}=4t$ ...(Tangent condition of parabola)
$\Rightarrow y=\frac{x}{t}+4t$
$\therefore y_0=4t$
$G\to (0,4t)$
Area of triangle = $\frac{1}{2}\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 1\\ x_3& y_3& 1\end{array}\right|=\frac{1}{2}\left|\begin{array}{ccc}0& 3& 1\\ 0& 4t& 1\\ 4t^2& 8t& 1\end{array}\right|$
$\Rightarrow A\left(t\right)=6t^2-8t^3$
$A^{\prime }\left(t\right)-12t-24t^2=0$
$\Rightarrow t=\frac{1}{2}$
$A\left(t\right)=6(\frac{1}{2}{)}^2-8(\frac{1}{2}{)}^3=\frac{3}{2}-1=\frac{1}{2}$
G comes out to be $(0,2)\Rightarrow y_1=2$
So, $F(1,4)$
Put F in $y=mx+3$ since it passes through it.
$\therefore m=1$