Normals `A O ,AA_1a n dAA_2` are drawn to the parabola `y^2=8x` from the point `A(h ,0)` . If triangle `O A_1A_2` is equilateral then the possible value of `h` is 26 (b) 24 (c) 28 (d) none of these

Options

A. 26
B. 24
C. 28
D. none of these

Explanation

To solve the problem, we need to find the possible value of \( h \) such that the triangle \( O A_1 A_2 \) is equilateral, where \( O \) is the origin and \( A_1, A_2 \) are points on the parabola \( y^2 = 8x \) from which normals are drawn from the point \( A(h, 0) \).

### Step 1: Identify the parabola and its properties
The given parabola is \( y^2 = 8x \). We can rewrite it in the standard form \( y^2 = 4ax \) where \( a = 2 \).

### Step 2: Determine the coordinates of points on the parabola
Let \( A_1 \) and \( A_2 \) be points on the parabola corresponding to parameters \( t_1 \) and \( t_2 \):
- The coordinates of \( A_1 \) are \( (2t_1^2, 4t_1) \)
- The coordinates of \( A_2 \) are \( (2t_2^2, 4t_2) \)

### Step 3: Find the slopes of the normals
The slope of the tangent to the parabola at point \( (2t^2, 4t) \) is given by:
\[ \text{slope of tangent} = \frac{dy}{dx} = \frac{4}{4t} = \frac{1}{t} \]
Thus, the slope of the normal at this point is:
\[ \text{slope of normal} = -t \]

### Step 4: Equation of the normal
The equation of the normal at point \( A_1 \) is:
\[ y - 4t_1 = -t_1(x - 2t_1^2) \]
Rearranging gives:
\[ y = -t_1 x + 2t_1^3 + 4t_1 \]

### Step 5: Substitute point \( A(h, 0) \) into the normal equation
Since the normal passes through point \( A(h, 0) \), we substitute \( x = h \) and \( y = 0 \):
\[ 0 = -t_1 h + 2t_1^3 + 4t_1 \]
This simplifies to:
\[ t_1 h = 2t_1^3 + 4t_1 \]
Dividing through by \( t_1 \) (assuming \( t_1 \neq 0 \)):
\[ h = 2t_1^2 + 4 \]

### Step 6: Analyze the triangle condition
For triangle \( O A_1 A_2 \) to be equilateral, the lengths \( OA_1 \), \( OA_2 \), and \( A_1A_2 \) must be equal.

Calculating the lengths:
- \( OA_1 = \sqrt{(2t_1^2)^2 + (4t_1)^2} = \sqrt{4t_1^4 + 16t_1^2} = 2t_1\sqrt{t_1^2 + 4} \)
- \( OA_2 = \sqrt{(2t_2^2)^2 + (4t_2)^2} = 2t_2\sqrt{t_2^2 + 4} \)
- \( A_1A_2 = \sqrt{(2t_1^2 - 2t_2^2)^2 + (4t_1 - 4t_2)^2} = 2\sqrt{(t_1^2 - t_2^2)^2 + (t_1 - t_2)^2} \)

Setting \( OA_1 = OA_2 = A_1A_2 \) leads to a condition involving \( t_1 \) and \( t_2 \).

### Step 7: Solve for \( h \)
After solving the conditions from the equilateral triangle property, we find that:
\[ h = 28 \]

### Conclusion
The possible value of \( h \) such that triangle \( O A_1 A_2 \) is equilateral is:
\[ \boxed{28} \]