find all the lines that pass through the point ` (1, 1)` and are tangent to the curve represent parametrically as `x = 2t-t^2` and `y = t + t^2`

Options

A. `(2sqrt(43))/(9)`
B. 2
C. 3
D. `(2sqrt(53))/(9)`

Explanation

To find all the lines that pass through the point (1, 1) and are tangent to the curve represented parametrically by \( x = 2t - t^2 \) and \( y = t + t^2 \), we will follow these steps:

### Step 1: Find the derivatives
First, we need to find the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

Given:
\[ x = 2t - t^2 \]
\[ y = t + t^2 \]

Calculating the derivatives:
\[ \frac{dx}{dt} = 2 - 2t \]
\[ \frac{dy}{dt} = 1 + 2t \]

### Step 2: Find the slope of the tangent line
The slope of the tangent line \( \frac{dy}{dx} \) can be found using the chain rule:
\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{1 + 2t}{2 - 2t} \]

### Step 3: Set up the equation of the tangent line
The equation of the tangent line at the point \( (x(t), y(t)) \) can be expressed using point-slope form:
\[ y - y(t) = \frac{dy}{dx} (x - x(t)) \]

### Step 4: Substitute the point (1, 1)
We want this line to pass through the point (1, 1). Therefore, substituting \( x = 1 \) and \( y = 1 \):
\[ 1 - (t + t^2) = \frac{1 + 2t}{2 - 2t} (1 - (2t - t^2)) \]

### Step 5: Simplify the equation
Substituting \( x(t) \) and \( y(t) \) into the equation:
\[ 1 - (t + t^2) = \frac{1 + 2t}{2 - 2t} (1 - (2t - t^2)) \]
This simplifies to:
\[ 1 - t - t^2 = \frac{1 + 2t}{2 - 2t} (t^2 - 2t + 1) \]

### Step 6: Cross-multiply and rearrange
Cross-multiplying gives:
\[ (1 - t - t^2)(2 - 2t) = (1 + 2t)(t^2 - 2t + 1) \]

Expanding both sides leads to a polynomial equation in \( t \):
\[ 2 - 2t - 2t^2 + 2t^2 + 2t^3 = t^2 + 2t^3 - 2t + 1 + 2t^2 \]

### Step 7: Solve the polynomial equation
Rearranging terms gives:
\[ 3t^2 - 4t + 1 = 0 \]
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6} \]
This results in:
\[ t = 1 \quad \text{and} \quad t = \frac{1}{3} \]

### Step 8: Find the points on the curve
For \( t = 1 \):
\[ x(1) = 2(1) - (1)^2 = 1, \quad y(1) = 1 + (1)^2 = 2 \quad \Rightarrow \quad (1, 2) \]

For \( t = \frac{1}{3} \):
\[ x\left(\frac{1}{3}\right) = 2\left(\frac{1}{3}\right) - \left(\frac{1}{3}\right)^2 = \frac{2}{3} - \frac{1}{9} = \frac{5}{9} \]
\[ y\left(\frac{1}{3}\right) = \frac{1}{3} + \left(\frac{1}{3}\right)^2 = \frac{1}{3} + \frac{1}{9} = \frac{4}{9} \quad \Rightarrow \quad \left(\frac{5}{9}, \frac{4}{9}\right) \]

### Step 9: Find the distance from (1, 1) to the points
Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For \( (1, 2) \):
\[ d_1 = \sqrt{(1 - 1)^2 + (2 - 1)^2} = \sqrt{0 + 1} = 1 \]
For \( \left(\frac{5}{9}, \frac{4}{9}\right) \):
\[ d_2 = \sqrt{\left(\frac{5}{9} - 1\right)^2 + \left(\frac{4}{9} - 1\right)^2} = \sqrt{\left(-\frac{4}{9}\right)^2 + \left(-\frac{5}{9}\right)^2} = \sqrt{\frac{16}{81} + \frac{25}{81}} = \sqrt{\frac{41}{81}} = \frac{\sqrt{41}}{9} \]

### Final Answer
The lines that pass through the point \( (1, 1) \) and are tangent to the curve are at the points \( (1, 2) \) and \( \left(\frac{5}{9}, \frac{4}{9}\right) \).