If the tangent at P (1, 1) on `y^2 = x (2-x)^2` meets the curve again at Q then the point Q is

Options

A. `(-1,2)`
B. `(9/4,3/8)`
C. `(4,4)`
D. None of these

Answer

To find the point Q where the tangent at point P (1, 1) on the curve \( y^2 = x(2-x)^2 \) meets the curve again, we will follow these steps:

### Step 1: Differentiate the Curve
We start with the equation of the curve:
\[ y^2 = x(2-x)^2 \]
To find the slope of the tangent at point P, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(x(2-x)^2) \]
Using implicit differentiation:
\[ 2y \frac{dy}{dx} = (2-x)^2 + x \cdot 2(2-x)(-1) \]
This simplifies to:
\[ 2y \frac{dy}{dx} = (2-x)^2 - 2x(2-x) \]

### Step 2: Evaluate the Derivative at Point P
Now, we substitute \( x = 1 \) and \( y = 1 \) into the derivative to find the slope at point P:
\[ 2(1) \frac{dy}{dx} = (2-1)^2 - 2(1)(2-1) \]
Calculating:
\[ 2 \frac{dy}{dx} = 1 - 2 = -1 \]
Thus, the slope \( m \) at point P is:
\[ \frac{dy}{dx} = -\frac{1}{2} \]

### Step 3: Write the Equation of the Tangent Line
Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = -\frac{1}{2} \), \( x_1 = 1 \), and \( y_1 = 1 \):
\[ y - 1 = -\frac{1}{2}(x - 1) \]
Rearranging gives:
\[ y = -\frac{1}{2}x + \frac{3}{2} \]

### Step 4: Substitute the Tangent Equation into the Curve
Next, we substitute \( y = -\frac{1}{2}x + \frac{3}{2} \) into the curve equation \( y^2 = x(2-x)^2 \):
\[ \left(-\frac{1}{2}x + \frac{3}{2}\right)^2 = x(2-x)^2 \]
Expanding both sides:
\[ \frac{1}{4}x^2 - \frac{3}{2}x + \frac{9}{4} = x(4 - 4x + x^2) \]
This simplifies to:
\[ \frac{1}{4}x^2 - \frac{3}{2}x + \frac{9}{4} = 4x - 4x^2 + x^3 \]

### Step 5: Rearranging the Equation
Rearranging gives us a polynomial equation:
\[ x^3 - \frac{17}{4}x^2 + \frac{11}{2}x - \frac{9}{4} = 0 \]
Multiplying through by 4 to eliminate the fractions:
\[ 4x^3 - 17x^2 + 22x - 9 = 0 \]

### Step 6: Factor the Polynomial
We can factor this polynomial to find the roots. Testing for rational roots, we find:
\[ (x - 1)^2(4x - 9) = 0 \]
This gives us roots \( x = 1 \) (double root) and \( x = \frac{9}{4} \).

### Step 7: Find Corresponding y-values
For \( x = \frac{9}{4} \):
Substituting back into the curve equation to find \( y \):
\[ y^2 = \frac{9}{4}(2 - \frac{9}{4})^2 = \frac{9}{4} \cdot \frac{1}{16} = \frac{9}{64} \]
Thus, \( y = \frac{3}{8} \) or \( y = -\frac{3}{8} \).

### Conclusion
The point Q where the tangent meets the curve again is:
\[ Q\left(\frac{9}{4}, \frac{3}{8}\right) \]