A normal at any point (x,y) to the curve y = f(x) cuts triangle of unit area with the axes, the equation of the curve is :

Options

A. `(x-1)/(y-1)=C`
B. `x/y=C`
C. xy=C
D. None of these

Answer

a

Explanation

The equation of the tangent at P(x,y) is
` Y - y = (dy)/(dx) (X-x)`
`((y-x((dy)/(dx)))/(1-(dy)/(dx)),(y-x(dy)/(dx))/(1-(dy)/(dx)))rArr (y-x(dy)/(dx))/(1-(dy)/(dx))=1`
`rArr" "y-x (dy)/(dx)=1-(dy)/(dx)`
`rArr" "(dx)/(x-1)=(dy)/(y-1)`
On integrating, we get
`(x-1)=C(y-1) rArr C=((x-1)/((y-1))`