Consider a curve y = y(x) in the first quadrant as shown in the...

Given that ${A}_{1} = 2{A}_{2}$

Now, from the graph

${A}_{1} + {A}_{2} = x \times y - 2 \times 4 = xy - 8$

$⇒ \dfrac{3}{2}{A}_{1} = xy - 8$

$⇒ {A}_{1} = \dfrac{2}{3}xy - \dfrac{16}{3}$

$⇒ {∫}_{4}^{x}f\left(x\right)dx = \dfrac{2}{3}xy - \dfrac{16}{3}$

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

$⇒ \dfrac{2}{3}x\dfrac{dy}{dx} = \dfrac{y}{3}$

$⇒ 2 ∫ \dfrac{dy}{y} = ∫ \dfrac{dx}{x}$

$⇒ 2\ln y = \ln x + lnc$

$⇒ {y}^{2} = cx$

As $f\left(4\right) = 2 ⇒ c = 1$

So ${y}^{2} = x$ is the given curve.

Given that the slope of normal $= - 6$

So the equation of normal will be

$y = \left(- 6x\right) - \dfrac{1}{2}$\left(- 6\right) - \dfrac{1}{4}{\left(- 6\right)}^{3}$</p><p>$⇒ y = - 6x + 3 + 54$</p><p>$⇒ y + 6x = 57$</p><p>Only \left(10, - 4\right)$ does not satisfy from the given options.