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 $\text{List I}$ with $\text{List II}$ and select the correct answer using the code given below the lists :

$\begin{array}{cccc}& \text{List I}& & \text{List II}\\ \text{(P)}& m=& \left(1\right)& \frac{1}{2}\\ \text{(Q)}& \text{Maximum area of triangle EFG is}& \left(2\right)& 4\\ \text{(R)}& y_0=& \left(3\right)& 2\\ \text{(S)}& y_1=& \left(4\right)& 1\end{array}$

Which of the following is correct combination

• A. $\left(P\right)\to \left(4\right),\left(Q\right)\to \left(1\right),\left(R\right)\to \left(2\right),\left(S\right)\to \left(3\right)$
• B. $\left(P\right)\to \left(3\right),\left(Q\right)\to \left(4\right),\left(R\right)\to \left(1\right),\left(S\right)\to \left(2\right)$
• C. $\left(P\right)\to \left(1\right),\left(Q\right)\to \left(3\right),\left(R\right)\to \left(2\right),\left(S\right)\to \left(4\right)$
• D. $\left(P\right)\to \left(1\right),\left(Q\right)\to \left(3\right),\left(R\right)\to \left(4\right),\left(S\right)\to \left(2\right)$

Answer 1

The correct option is A $\left(P\right)\to \left(4\right),\left(Q\right)\to \left(1\right),\left(R\right)\to \left(2\right),\left(S\right)\to \left(3\right)$



Area of $△EFG$
$A\left(t\right)=\frac{1}{2}\left|\begin{array}{ccc}1& 1& 1\\ 0& 0& 4t^2\\ 3& 4t& 8t\end{array}\right|=12t^2-48t^3\Rightarrow A^{\prime }\left(t\right)=24t(-2t+1)$
$t=0$ (Not possible )
Hence, $t=\frac{1}{2}$

Points: $E(0,3),F(1,4),G(0,2)$
Hence, $y_0=4,y_1=2$
Slope $=m=\frac{4-3}{1-0}=1$
Area $=\frac{1}{2}\left|\begin{array}{ccc}0& 3& 1\\ 0& 2& 1\\ 1& 4& 1\end{array}\right|=\frac{1}{2}$ unit square