The normal to the curve y(x-2)(x-3)=x+6 at the point where the curve…

The equation of the curve is

$y(x-2)(x-3)=x+6$              ......(i)

It intersects y-axis at x = 0. Putting x = 0 in (i), we obtain y = 1. So, the point where (i) cuts y-axis has the coordinates (0, 1).

From (i), we obtain

$y=\frac{x+6}{x^2-5 x+6}$

Differentiating with respect to x, we obtain

$\frac{d y}{d x}=\frac{\left(x^2-5 x+6\right)-(x+6)(2 x-5)}{\left(x^2-5 x+6\right)^2}$

∴  $\left(\frac{d y}{d x}\right)_{(0,1)}=\frac{6-(-30)}{36}=1$

The equation of the normal at (0, 1) is

$y-1=-1(x-0)$ or, $x+y-1=0$

Clearly, it passes through $\left(\frac{1}{2}, \frac{1}{2}\right)$.