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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

$(-\frac{1}{2},-\frac{1}{2})$

$(\frac{1}{2},\frac{1}{2})$

$(\frac{1}{2},-\frac{1}{3})$

$(\frac{1}{2},\frac{1}{3})$

Option B

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 $(\frac{1}{2},\frac{1}{2})$

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