Discussion Forum

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

Answer:

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 $(\frac{1}{2},\frac{1}{2})$