Answer:
Option B
Explanation:
Given curve is
y(x-2)(x-3)=x+6 .........(i)
put x=0 in Eq.(i) , we get
y(-2)(-3)=6\Rightarrow y=1
So, point of intersection is (0,1)
Now, y= \frac{x+6}{(x-2)(x-3)}
\Rightarrow \frac{\text{d}y}{\text{d}x}= \frac{1(x-2)(x-3)-(x+6)(x-3+x-2)}{(x-2)^{2}(x-3)^{2}}
=(\frac{\text{d}y}{\text{d}x})_{(0,1)}=\frac{6+30}{4\times9}=\frac{36}{36}=1
\therefore Equation of normal at (0,1) is given by
y-1 =\frac{-1}{1}(x-0)
\Rightarrow x+y-1=0
which passes through the point (\frac{1}{2},\frac{1}{2})