Find the intersection S of the lines `x-ty+t^(2)=0, tx+y-t^(3)-2t=0`.
Show that S lies on the curve whose equation is `y^(2)=4x`. Sketch this curve.

Answer

To solve the problem, we need to find the intersection point \( S \) of the two lines given by the equations:

1. \( x - ty + t^2 = 0 \) (Equation 1)
2. \( tx + y - t^3 - 2t = 0 \) (Equation 2)

### Step 1: Rearranging the equations
First, we can rearrange both equations to express \( y \) in terms of \( x \) and \( t \).

From Equation 1:
\[ x - ty + t^2 = 0 \implies ty = x + t^2 \implies y = \frac{x + t^2}{t} \]

From Equation 2:
\[ tx + y - t^3 - 2t = 0 \implies y = t^3 + 2t - tx \]

### Step 2: Setting the equations for \( y \) equal
Now we have two expressions for \( y \):
\[ \frac{x + t^2}{t} = t^3 + 2t - tx \]

### Step 3: Multiplying through by \( t \)
To eliminate the fraction, multiply through by \( t \):
\[ x + t^2 = t(t^3 + 2t - tx) \]
Expanding the right side:
\[ x + t^2 = t^4 + 2t^2 - t^2x \]

### Step 4: Rearranging to isolate \( x \)
Rearranging gives:
\[ x + t^2x = t^4 + 2t^2 - t^2 \]
\[ x(1 + t^2) = t^4 + t^2 \]
\[ x = \frac{t^4 + t^2}{1 + t^2} \]

### Step 5: Simplifying \( x \)
Factoring out \( t^2 \):
\[ x = \frac{t^2(t^2 + 1)}{1 + t^2} = t^2 \]

### Step 6: Finding \( y \)
Now substituting \( x = t^2 \) back into one of the original equations to find \( y \). Using Equation 1:
\[ t^2 - ty + t^2 = 0 \implies -ty + 2t^2 = 0 \implies ty = 2t^2 \implies y = 2t \]

### Step 7: Conclusion for point \( S \)
Thus, the intersection point \( S \) is:
\[ S = (t^2, 2t) \]

### Step 8: Verifying if \( S \) lies on the curve \( y^2 = 4x \)
We need to verify if the point \( S(t^2, 2t) \) lies on the curve defined by \( y^2 = 4x \):
\[ y^2 = (2t)^2 = 4t^2 \]
\[ 4x = 4(t^2) \]
Since \( y^2 = 4x \), the point \( S \) lies on the curve.

### Step 9: Sketching the curve
The equation \( y^2 = 4x \) represents a parabola that opens to the right. The vertex of the parabola is at the origin (0,0).

### Summary of Steps:
1. Rearranged both equations to express \( y \).
2. Set the two expressions for \( y \) equal.
3. Eliminated the fraction by multiplying through by \( t \).
4. Rearranged to isolate \( x \).
5. Found \( x = t^2 \).
6. Substituted \( x \) back to find \( y = 2t \).
7. Verified that \( S(t^2, 2t) \) lies on the curve \( y^2 = 4x \).
8. Recognized the curve as a rightward-opening parabola.