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 : `(1/2,-1/3)` (2) `(1/2,1/3)` (3) `(-1/2,-1/2)` (4) `(1/2,1/2)`

Options

A. `(-(1)/(2), -(1)/(2))`
B. `((1)/(2), (1)/(2))`
C. `((1)/(2), - (1)/(3))`
D. `((1)/(2), (1)/(3))`

Explanation

Given curve is
` y (x - 2) ( x- 3) = x + 6" " `…(i)
Put `x =0` in Eq. (i), we get
`y (-2)(-3) = 6 rArr y = 1`
So, point of intersection is `(0, 1)`.
Now, `y = (x + 6)/((x - 2)(x - 3))`
`rArr (dy)/(dx) = (1(x- 2)(x-3) - (x+ 6)( x- 3 + x - 2))/((x - 2)^(2)( x - 3)^(2))`
`rArr ((dy)/(dx))_("("0,1")")= (6+ 30)/( 4xx9) = ( 36)/(36) = 1 `
`therefore` Equation of normal at `(0, 1)` is given by
`y - 1 = (-1)/(1) (x-0)`
`rArr x + y - 1 =0`
which passes through the point `((1)/(2), (1)/(2))`.