Consider a curve y = y x in the first quadrant as shown in the figure. Let the area A 1 is twice the area A…

Given that $A_1=2A_2$

Now, from the graph

$A_1+A_2=x\times y-2\times 4=xy-8$

$\Rightarrow \frac{3}{2} A_1=xy-8$

$\Rightarrow A_1=\frac{2}{3}xy-\frac{16}{3}$

$\Rightarrow {\int }_4^{x}f\left(x\right)dx=\frac{2}{3}xy-\frac{16}{3}$

$\Rightarrow f\left(x\right)=\frac{2}{3}\left(x\frac{dy}{dx}+y\right)$

$\Rightarrow \frac{2}{3}x\frac{dy}{dx}=\frac{y}{3}$

$\Rightarrow 2\int \frac{dy}{y}=\int \frac{dx}{x}$

$\Rightarrow 2\ln y=\ln x+\ln c$

$\Rightarrow y^2=cx$ 

As $f\left(4\right)=2\Rightarrow c=1$

So $y^2=x$ is the given curve.

Given that the slope of normal $=-6$

So the equation of normal will be 

$y=\left(-6x\right)-\frac{1}{2}\left(-6\right)-\frac{1}{4}{\left(-6\right)}^3$

$\Rightarrow y=-6x+3+54$

$\Rightarrow y+6x=57$

Only $\left(10,-4\right)$ does not satisfy from the given options.