if the tangent to the parabola `y=x(2-x)` at the point (1,1) intersects the parabola at P. find the co-ordinate of P.

Options

A. (1,2)
B. (1,1)
C. (3,-2)
D. (-1,-4)

Answer

To find the coordinates of point P where the tangent to the parabola \( y = x(2 - x) \) at the point (1, 1) intersects the parabola again, we can follow these steps:

### Step 1: Find the equation of the parabola
The given parabola can be rewritten as:
\[ y = 2x - x^2 \]

### Step 2: Find the derivative to get the slope of the tangent
To find the slope of the tangent at the point (1, 1), we need to differentiate the equation of the parabola:
\[ \frac{dy}{dx} = 2 - 2x \]
Now, substituting \( x = 1 \):
\[ \frac{dy}{dx} \bigg|_{x=1} = 2 - 2(1) = 0 \]
So, the slope of the tangent at (1, 1) is 0.

### Step 3: Write the equation of the tangent line
Since the slope is 0, the equation of the tangent line at (1, 1) is a horizontal line:
\[ y = 1 \]

### Step 4: Find the points of intersection of the tangent line and the parabola
Now, we need to find the points where this tangent line \( y = 1 \) intersects the parabola \( y = 2x - x^2 \):
\[ 1 = 2x - x^2 \]
Rearranging gives:
\[ x^2 - 2x + 1 = 0 \]
This can be factored as:
\[ (x - 1)^2 = 0 \]
Thus, we have:
\[ x = 1 \]

### Step 5: Find the y-coordinate for the intersection point
Substituting \( x = 1 \) back into the equation of the parabola to find the corresponding y-coordinate:
\[ y = 2(1) - (1)^2 = 2 - 1 = 1 \]
So, the point of intersection is (1, 1).

### Step 6: Identify the other intersection point
Since the tangent line intersects the parabola at (1, 1) and is a horizontal line, it does not intersect the parabola at any other point. Therefore, point P is the same as the point of tangency.

### Conclusion
The coordinates of point P are:
\[ \text{P} = (1, 1) \]

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