The normal to the curve, \(y(x-2)(x-3)=x+6\) at the point, where the curve intersects the Y-axis, passes through the point

Given equation of curve
\(\begin{aligned}
& y(x-2)(x-3)=x+6 \\
& \Rightarrow y=\frac{x+6}{(x-2)(x-3)} \\
& \Rightarrow y=\frac{x+6}{x^2-5 x+6}
\end{aligned}\)
Differentiating w.r.t. \(x\), we get
\(\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\left(x^2-5 x+6\right)(1)-(x+6)(2 x-5)}{\left(x^2-5 x+6\right)^2} \\
& =\frac{x^2-5 x+6-(x+6)(2 x-5)}{\left(x^2-5 x+6\right)^2} \\
& \therefore \quad \frac{\mathrm{~d} y}{\mathrm{~d} x}=\frac{-x^2-12 x+36}{\left(x^2-5 x+6\right)^2} \\
& \text { At Y-axis, } x=0
\end{aligned}\)
\(\begin{aligned} \therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x_{\mathrm{at} x=0}} & =\frac{-(0)^2-12(0)+36}{\left(0^2-5(0)+6\right)^2} \\ & =\frac{36}{36} \\ & =1\end{aligned}\)
The equation of normal is
\(y-1=-1(x-0)\)
i.e., \(x+y=1\)
Option (B) i.e.,
\(\therefore \quad\left(\frac{1}{2}, \frac{1}{2}\right)\) satisfies above equation
\(\therefore \quad\) Normal passes through \(\left(\frac{1}{2}, \frac{1}{2}\right)\).