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

Options

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

Answer

To solve the problem, we need to find the point \( P \) where the tangent at the point \( (1, 1) \) on the curve \( y^2 = x(2 - x)^2 \) meets the curve again. Here are the steps:

### Step 1: Find the equation of the curve
The given curve is:
\[ y^2 = x(2 - x)^2 \]

### Step 2: Differentiate the curve
To find the slope of the tangent line at any point on the curve, we differentiate both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = (2 - x)^2 + x \cdot 2(2 - x)(-1) \]
Simplifying this:
\[ 2y \frac{dy}{dx} = (2 - x)^2 - 2x(2 - x) \]
\[ = (2 - x)^2 - 4x + 2x^2 \]
\[ = 4 - 4x + x^2 - 4x + 2x^2 = 4 - 8x + 3x^2 \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{4 - 8x + 3x^2}{2y} \]

### Step 3: Evaluate the derivative at the point (1, 1)
Substituting \( x = 1 \) and \( y = 1 \):
\[ \frac{dy}{dx} = \frac{4 - 8(1) + 3(1)^2}{2(1)} = \frac{4 - 8 + 3}{2} = \frac{-1}{2} \]

### Step 4: Write the equation of the tangent line
Using the point-slope form of the equation of the tangent line:
\[ y - y_1 = m(x - x_1) \]
where \( (x_1, y_1) = (1, 1) \) and \( m = -\frac{1}{2} \):
\[ y - 1 = -\frac{1}{2}(x - 1) \]
Rearranging gives:
\[ y - 1 = -\frac{1}{2}x + \frac{1}{2} \]
\[ y = -\frac{1}{2}x + \frac{3}{2} \]

### Step 5: Substitute the tangent equation into the curve equation
Now, 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:
\[ \frac{1}{4}x^2 - \frac{3}{2}x + \frac{9}{4} = x(4 - 4x + x^2) \]
\[ \frac{1}{4}x^2 - \frac{3}{2}x + \frac{9}{4} = 4x - 4x^2 + x^3 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ x^3 - \frac{17}{4}x^2 + \frac{11}{2}x - \frac{9}{4} = 0 \]
Multiplying through by 4 to eliminate the fractions:
\[ 4x^3 - 17x^2 + 22x - 9 = 0 \]

### Step 7: Factor the cubic equation
Using synthetic division or the Rational Root Theorem, we find that \( x = 1 \) is a root. Thus, we can factor:
\[ (x - 1)(4x^2 - 13x + 9) = 0 \]
Solving \( 4x^2 - 13x + 9 = 0 \) using the quadratic formula:
\[ x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 4 \cdot 9}}{2 \cdot 4} \]
\[ x = \frac{13 \pm \sqrt{169 - 144}}{8} = \frac{13 \pm 5}{8} \]
Thus, we have:
\[ x = \frac{18}{8} = \frac{9}{4} \quad \text{and} \quad x = 1 \]

### Step 8: Find corresponding \( y \) values
Now, substituting \( x = \frac{9}{4} \) back into the tangent line equation to find \( y \):
\[ y = -\frac{1}{2}\left(\frac{9}{4}\right) + \frac{3}{2} = -\frac{9}{8} + \frac{12}{8} = \frac{3}{8} \]

### Final Answer
Thus, the point \( P \) where the tangent meets the curve again is:
\[ P = \left(\frac{9}{4}, \frac{3}{8}\right) \]