JEE Main · 2017 · 02 Apr 2017 · Offline · Differentiation

At the $y$-axis $x=0$, so $y(0-2)(0-3)=0+6$ gives $6y=6$, that is the point $(0,1)$. Differentiating implicitly, $y(2x-5)+y^{\prime }(x^2-5x+6)=1$; at $(0,1)$ this reads $-5+6y^{\prime }=1$, so $y^{\prime }=1$. The tangent slope is $1$, hence the normal slope is $-1$, and the normal is $y-1=-(x-0)$, i.e. $y=1-x$. The point $\left(\frac{1}{2},\frac{1}{2}\right)$ satisfies it, so the normal passes through $\left(\frac{1}{2},\frac{1}{2}\right)$.