The point of intersection of the tangents drawn to the curve `x^(2)y=1-y` at the points where it is meet by he cuver xy=1-y is given by :

Options

A. `(0,-1)`
B. `(1,1)`
C. `(0,1)`
D. none of these

Answer

C

Explanation

Solving the two equations, we get
` x^(2)y =xy rArr xy(x-1)=0 rArr x=0, y=0, x=1 `
Since y = 0 does not satisfy the two equations. So, we neglect it.
Putting x = 0 in the either equation, we get x = 1.
Now, putting x=1 in one of the two equations we obtain ` y= 1//2. ` Thus, the two curves intersect at (0,1) and `(1,1//2).`
Now,
` x^(2)y=1-y rArr x^(2)(dy)/(dx) + 2xy = -(dy)/(dx) rArr (dy)/(dx) =-(2xy)/(x^(2)+1) `
` rArr ((dy)/(dx))_((0","1)) =0 " and " ((dy)/(dx))_((1"," 1//2))=-(1)/(2)`
The equations of the required tangents are
` y-1 =0(x-0) " and " y-(1)/(2)=(-1)/(2)(x-1) `
` rArr y=1 " and " x+2y-2=0 `
These two tangents intersect at (0,1)