Let `C : y =x^(2) -3, D : y = kx^(2)` be two parabolas and
`L_(1)` : `x= a , L_(2) : x = 1 (a ne 0)` be two straight lines.
IF the line `L_(1)` meets the parabola C at a point B on the line `L_(2)`, other than A, then a may be equal to

Options

A. `A) -3`
B. B)`-2`
C. C) 2
D. D) None of these

Answer

To solve the problem, we need to find the value of \( a \) such that the line \( L_1: x = a \) intersects the parabola \( C: y = x^2 - 3 \) at a point \( B \) which lies on the line \( L_2: x = 1 \).

### Step-by-Step Solution:

1. **Identify the Points of Intersection**:
- The line \( L_1 \) intersects the parabola \( C \) at the point \( (a, y_1) \).
- The line \( L_2 \) intersects the parabola \( C \) at the point \( (1, y_2) \).

2. **Calculate \( y_2 \)**:
- Substitute \( x = 1 \) into the equation of parabola \( C \):
\[ y_2 = 1^2 - 3 = 1 - 3 = -2 \]
- Thus, the coordinates of point \( B \) are \( (1, -2) \).

3. **Find the Coordinates of Point \( A \)**:
- The point \( A \) has coordinates \( (a, y_1) \).
- Substitute \( x = a \) into the equation of parabola \( C \):
\[ y_1 = a^2 - 3 \]
- Therefore, the coordinates of point \( A \) are \( (a, a^2 - 3) \).

4. **Find the Slope of the Tangent at Point \( A \)**:
- The derivative of \( y = x^2 - 3 \) gives the slope of the tangent:
\[ \frac{dy}{dx} = 2x \]
- At point \( A \) where \( x = a \):
\[ \text{slope at } A = 2a \]

5. **Find the Slope of the Line \( AB \)**:
- The slope of the line \( AB \) connecting points \( A(a, a^2 - 3) \) and \( B(1, -2) \) is given by:
\[ \text{slope of } AB = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - (a^2 - 3)}{1 - a} = \frac{-2 - a^2 + 3}{1 - a} = \frac{1 - a^2}{1 - a} \]

6. **Set the Slopes Equal**:
- For the tangent at point \( A \) to pass through point \( B \), the slopes must be equal:
\[ 2a = \frac{1 - a^2}{1 - a} \]

7. **Cross Multiply and Simplify**:
- Cross multiplying gives:
\[ 2a(1 - a) = 1 - a^2 \]
- Expanding and rearranging:
\[ 2a - 2a^2 = 1 - a^2 \]
\[ 2a^2 - a + 1 = 0 \]

8. **Solve the Quadratic Equation**:
- Rearranging gives:
\[ a^2 - a + 1 = 0 \]
- Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ a = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} \]

9. **Conclusion**:
- Since the discriminant is negative, there are no real solutions for \( a \). Hence, the problem might have constraints or conditions that were not specified.