Find the point of intersection of the tangents drawn to the curve `x^2y=1-y` at the points where it is intersected by the curve `x y=1-ydot`

Options

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

Answer

To solve the problem, we need to find the point of intersection of the tangents drawn to the curve \( x^2y = 1 - y \) at the points where it intersects with the curve \( xy = 1 - y \).

### Step 1: Find the intersection points of the curves

We have two curves:
1. \( x^2y = 1 - y \)
2. \( xy = 1 - y \)

Rearranging the first curve gives:
\[ y(x^2 + 1) = 1 \implies y = \frac{1}{x^2 + 1} \]

Rearranging the second curve gives:
\[ y(x + 1) = 1 \implies y = \frac{1}{x + 1} \]

Now, we need to equate the two expressions for \( y \):
\[ \frac{1}{x^2 + 1} = \frac{1}{x + 1} \]

### Step 2: Cross-multiply to eliminate the fractions

Cross-multiplying gives:
\[ x + 1 = x^2 + 1 \]

### Step 3: Rearrange the equation

Rearranging the equation:
\[ x^2 - x = 0 \]

### Step 4: Factor the equation

Factoring gives:
\[ x(x - 1) = 0 \]

### Step 5: Solve for \( x \)

This gives us two solutions:
\[ x = 0 \quad \text{or} \quad x = 1 \]

### Step 6: Find corresponding \( y \) values

Now we find the corresponding \( y \) values for each \( x \):
1. For \( x = 0 \):
\[ y = \frac{1}{0 + 1} = 1 \]
So, one intersection point is \( (0, 1) \).

2. For \( x = 1 \):
\[ y = \frac{1}{1 + 1} = \frac{1}{2} \]
So, another intersection point is \( (1, \frac{1}{2}) \).

### Step 7: Find the tangents at the intersection points

Next, we need to find the tangents to the curve \( x^2y = 1 - y \) at these points.

#### For point \( (0, 1) \):

1. Differentiate \( x^2y = 1 - y \) implicitly:
\[ 2xy + x^2 \frac{dy}{dx} = -\frac{dy}{dx} \]
Rearranging gives:
\[ (x^2 + 1) \frac{dy}{dx} = -2xy \implies \frac{dy}{dx} = \frac{-2xy}{x^2 + 1} \]

2. Substitute \( (0, 1) \):
\[ \frac{dy}{dx} = \frac{-2(0)(1)}{0^2 + 1} = 0 \]
The tangent line at \( (0, 1) \) is horizontal: \( y = 1 \).

#### For point \( (1, \frac{1}{2}) \):

1. Substitute \( (1, \frac{1}{2}) \):
\[ \frac{dy}{dx} = \frac{-2(1)(\frac{1}{2})}{1^2 + 1} = \frac{-1}{2} \]
The equation of the tangent line using point-slope form:
\[ y - \frac{1}{2} = -\frac{1}{2}(x - 1) \]
Simplifying gives:
\[ y = -\frac{1}{2}x + 1 \]

### Step 8: Find the intersection of the tangents

Now we need to find the intersection of the two tangent lines:
1. \( y = 1 \)
2. \( y = -\frac{1}{2}x + 1 \)

Setting them equal:
\[ 1 = -\frac{1}{2}x + 1 \]
This simplifies to:
\[ 0 = -\frac{1}{2}x \implies x = 0 \]

Substituting \( x = 0 \) back into either equation gives \( y = 1 \).

### Final Result

Thus, the point of intersection of the tangents is:
\[ \boxed{(0, 1)} \]