Consider a curve y = y(x) in the first quadrant as shown in the figure. Let the area A₁ is twice the area A₂. Then the normal to the curve perpendicular to the line 2x – 12y = 15 does NOT pass through the point.

Given that ${\mathrm{A}}_1=2{\mathrm{A}}_2$

from the graph ${\mathrm{A}}_1+{\mathrm{A}}_2=\mathrm{xy}-8$

$\begin{array}{l}\Rightarrow \frac{3}{2}{\mathrm{A}}_1=\mathrm{xy}-8\\ \Rightarrow {\mathrm{A}}_1=\frac{2}{3}\mathrm{xy}-\frac{16}{3}\\ \Rightarrow {\int }_4^{\mathrm{x}} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\frac{2}{3}\mathrm{xy}-\frac{16}{3}\\ \Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{2}{3}\left(\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}\right)\\ \Rightarrow \frac{2}{3}\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{3}\\ \Rightarrow 2\int \frac{\mathrm{dy}}{\mathrm{y}}=\int \frac{\mathrm{dx}}{\mathrm{x}}\\ \Rightarrow 2\mathrm{\ell ny}=\mathrm{\ell nx}+\mathrm{\ell nc}\\ \Rightarrow {\mathrm{y}}^2=\mathrm{cx}\end{array}$

As f(4) = 2    $\Rightarrow$ c = 1

so y² = x

slope of normal = –6

$\begin{array}{l}\mathrm{y}=-6\left(\mathrm{x}\right)-\frac{1}{2}(-6)-\frac{1}{4}(-6{)}^3\\ \Rightarrow \mathrm{y}=-6\mathrm{x}+3+54\\ \Rightarrow \mathrm{y}+6\mathrm{x}=57\end{array}$

Now check options and (C) will not satisfy. the equation of normal