The point of intersection of the curve whose parametrix equations are `x=t^(2)+1, y=2t" and " x=2s, y=2/s,` is given by

Options

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

Answer

To find the point of intersection of the two curves given by their parametric equations, we will follow these steps:

### Step 1: Write down the parametric equations
The first curve is given by:
- \( x = t^2 + 1 \)
- \( y = 2t \)

The second curve is given by:
- \( x = 2s \)
- \( y = \frac{2}{s} \)

### Step 2: Express \( t \) in terms of \( y \) from the first curve
From the equation \( y = 2t \), we can express \( t \) as:
\[ t = \frac{y}{2} \]

### Step 3: Substitute \( t \) into the equation for \( x \)
Now substitute \( t = \frac{y}{2} \) into the equation for \( x \):
\[ x = \left(\frac{y}{2}\right)^2 + 1 \]
This simplifies to:
\[ x = \frac{y^2}{4} + 1 \]

### Step 4: Rearrange the equation for the first curve
Rearranging gives:
\[ 4x = y^2 + 4 \]
Thus, the equation of the first curve is:
\[ 4x - y^2 - 4 = 0 \quad \text{(1)} \]

### Step 5: Express \( s \) in terms of \( y \) from the second curve
From the equation \( y = \frac{2}{s} \), we can express \( s \) as:
\[ s = \frac{2}{y} \]

### Step 6: Substitute \( s \) into the equation for \( x \)
Now substitute \( s = \frac{2}{y} \) into the equation for \( x \):
\[ x = 2s = 2 \left(\frac{2}{y}\right) = \frac{4}{y} \]

### Step 7: Rearrange the equation for the second curve
Thus, the equation of the second curve is:
\[ xy = 4 \quad \text{(2)} \]

### Step 8: Solve the system of equations
Now we have two equations:
1. \( 4x - y^2 - 4 = 0 \)
2. \( xy = 4 \)

From equation (2), we can express \( x \) in terms of \( y \):
\[ x = \frac{4}{y} \]

### Step 9: Substitute \( x \) into the first equation
Substituting \( x = \frac{4}{y} \) into equation (1):
\[ 4\left(\frac{4}{y}\right) - y^2 - 4 = 0 \]
This simplifies to:
\[ \frac{16}{y} - y^2 - 4 = 0 \]

### Step 10: Clear the fraction
Multiply through by \( y \) (assuming \( y \neq 0 \)):
\[ 16 - y^3 - 4y = 0 \]
Rearranging gives:
\[ y^3 + 4y - 16 = 0 \]

### Step 11: Find the roots of the polynomial
We can use the Rational Root Theorem or trial and error to find a root. Testing \( y = 2 \):
\[ 2^3 + 4(2) - 16 = 8 + 8 - 16 = 0 \]
Thus, \( y = 2 \) is a root.

### Step 12: Find the corresponding \( x \)
Now substitute \( y = 2 \) back into equation (2) to find \( x \):
\[ x(2) = 4 \implies x = 2 \]

### Conclusion
The point of intersection is:
\[ (x, y) = (2, 2) \]