Let points `A_(1), A_(2) and A_(3)` lie on the parabola `y^(2)=8x.` If `triangle A_(1)A_(2)A_(3)` is an equilateral triangle and normals at points `A_(1), A_(2) and A_(3)` on this parabola meet at the point (h, 0). Then the value of h I s

Options

A. 24
B. 26
C. 38
D. 28

Answer

To solve the problem, we need to find the value of \( h \) where the normals at points \( A_1, A_2, \) and \( A_3 \) on the parabola \( y^2 = 8x \) meet at the point \( (h, 0) \) and form an equilateral triangle.

### Step-by-Step Solution:

1. **Understand the Parabola**: The given parabola is \( y^2 = 8x \). This can be rewritten in standard form \( y^2 = 4ax \) where \( a = 2 \).

2. **Parametric Representation**: The points on the parabola can be represented parametrically as:
- \( A_1(t_1) = (t_1^2, 2 \cdot 2 \cdot t_1) = (t_1^2, 4t_1) \)
- \( A_2(t_2) = (t_2^2, 2 \cdot 2 \cdot t_2) = (t_2^2, 4t_2) \)
- \( A_3(t_3) = (t_3^2, 2 \cdot 2 \cdot t_3) = (t_3^2, 4t_3) \)

3. **Equilateral Triangle Condition**: For \( A_1, A_2, A_3 \) to form an equilateral triangle, the angles between the normals at these points must be \( 60^\circ \).

4. **Equation of the Normal**: The equation of the normal to the parabola at point \( (x_0, y_0) \) is given by:
\[ y - y_0 = -\frac{y_0}{4}(x - x_0) \]
For our points:
- At \( A_1(t_1) \): \( y - 4t_1 = -\frac{4t_1}{4}(x - t_1^2) \) simplifies to \( y = -t_1(x - t_1^2) + 4t_1 \).
- At \( A_2(t_2) \): \( y = -t_2(x - t_2^2) + 4t_2 \).
- At \( A_3(t_3) \): \( y = -t_3(x - t_3^2) + 4t_3 \).

5. **Finding the Intersection of Normals**: Since all normals meet at \( (h, 0) \), we can substitute \( y = 0 \) into the normal equations:
\[ 0 = -t_1(h - t_1^2) + 4t_1 \implies h = t_1^2 - 4 \]
\[ 0 = -t_2(h - t_2^2) + 4t_2 \implies h = t_2^2 - 4 \]
\[ 0 = -t_3(h - t_3^2) + 4t_3 \implies h = t_3^2 - 4 \]

6. **Equating the Values of \( h \)**: Since \( h \) is the same for all three points, we have:
\[ t_1^2 - 4 = t_2^2 - 4 = t_3^2 - 4 \]
This implies \( t_1^2 = t_2^2 = t_3^2 \).

7. **Using the Properties of Equilateral Triangle**: The slopes of the normals at \( A_1, A_2, A_3 \) must satisfy the condition for an equilateral triangle. The angles between the normals can be derived from the slopes.

8. **Finding \( t \)**: From the geometry of the equilateral triangle, we can derive that:
\[ t_1 = \sqrt{12}, \quad t_2 = -\sqrt{12}, \quad t_3 = 0 \]

9. **Calculating \( h \)**: Using \( t_1 = 2\sqrt{3} \):
\[ h = t_1^2 - 4 = 12 - 4 = 8 \]

Thus, the value of \( h \) is:

\[ \boxed{8} \]