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)`(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))`

Answer

B

Explanation

The equation of the curve is
`y(x-2)(x-3)=x+6 " " ` …(i)
It intersects y-axis at x = 0. Putting x = 0 in (i) , we obtain y = 1. So, the point where (i) cuts y-axis has the coordinates (0, 1).
Form (i), we obtain
`y=(x+6)/(x^(2)-5x+6)`
Differentiating with respect to x, we obtain
`(dy)/(dx) = ((x^(2)-5x+6)-(x+6)(2x-5))/((x^(2)-5x+6)^(2))`
`therefore ((dy)/(dx))_((0","1))=(6-(-30))/(36) =1`
The equation of the normal at (0, 1) is
`y-1= -1(x-0) or, x+y-1=0`
Clearly, it passes through `((1)/(2), (1)/(2))`.