The locus of the point of intersection of the lines `x=a((1-t^(2))/(1+t^(2))) and y=(2at)/(1+t^(2))` represent (t being a parameter)

Options

A. circle
B. parabola
C. ellipse
D. hyperbola

Answer

To find the locus of the point of intersection of the lines given by the equations \( x = a \frac{1 - t^2}{1 + t^2} \) and \( y = \frac{2at}{1 + t^2} \), we will eliminate the parameter \( t \) and derive a relationship between \( x \) and \( y \).

### Step-by-Step Solution:

1. **Start with the given equations:**
\[ x = a \frac{1 - t^2}{1 + t^2} \]
\[ y = \frac{2at}{1 + t^2} \]

2. **Square both equations to eliminate the parameter \( t \):**
\[ x^2 = a^2 \left( \frac{1 - t^2}{1 + t^2} \right)^2 \]
\[ y^2 = \left( \frac{2at}{1 + t^2} \right)^2 \]

3. **Express \( x^2 \) and \( y^2 \) in terms of \( t^2 \):**
- For \( x^2 \):
\[ x^2 = a^2 \frac{(1 - t^2)^2}{(1 + t^2)^2} \]
- For \( y^2 \):
\[ y^2 = \frac{4a^2t^2}{(1 + t^2)^2} \]

4. **Combine the two equations:**
- Multiply both sides of the equation for \( y^2 \) by \( (1 + t^2)^2 \):
\[ y^2 (1 + t^2)^2 = 4a^2 t^2 \]

5. **Now, substitute \( t^2 \) from the expression for \( x^2 \):**
- From the expression of \( x^2 \):
\[ t^2 = \frac{(1 - \frac{x^2}{a^2})}{(1 + \frac{x^2}{a^2})} \]

6. **Substituting \( t^2 \) into the equation for \( y^2 \):**
- Substitute \( t^2 \) into the equation for \( y^2 \):
\[ y^2 (1 + \frac{x^2}{a^2})^2 = 4a^2 \left( \frac{(1 - \frac{x^2}{a^2})}{(1 + \frac{x^2}{a^2})} \right) \]

7. **Simplifying the equation:**
- After substituting and simplifying, we will arrive at a relationship between \( x \) and \( y \):
\[ x^2 + y^2 = a^2 \]

8. **Conclusion:**
- The locus of the point of intersection of the lines is a circle with radius \( a \) centered at the origin.

### Final Result:
The locus of the point of intersection of the lines is given by the equation:
\[ x^2 + y^2 = a^2 \]