The curve given by `x + y = e^(xy)` has a tangent parallel to the y-axis at the point

Options

A. `(1,0)`
B. At no point
C. `(0,1)`
D. `(0,0)`

Answer

A

Explanation

Given equation is
`y-e^(xy)+x=0`
`rArr e^(xy)=x+y`
Taking log on both sides, we get
`log(x+y)=xy`
On differentiating, we get
`(1)/(x+y) (1+(dy)/(dx))=x(dy)/(dx)+y`
`rArr (dy)/(dx)((1)/(y+x)-x)=y-(1)/(y+x)`
`rArr (dy)/(dx)= (y(y+x)-1)/(1-x(y+x))`
Since, the curve has vertical tangent
i.e. `" "(dy)/(dx)=oo`
`rArr 1-x(x+y)=0`
which is satisfied by the point (1, 0).