If from a point A, two tangents are drawn to parabola `y^2 = 4ax` are normal to parabola `x^2 = 4by`, then

Options

A. `a^(2)geb^(2)`
B. `a^(2)ge4b^(2)`
C. `a^(2)ge8b^(2)`
D. `8a^(2)geb^(2)`

Explanation

To solve the problem, we need to derive the relationship between the parameters \( a \) and \( b \) given the conditions of the tangents and normals to the respective parabolas.

### Step-by-step Solution:

1. **Equation of the Tangent to the Parabola \( y^2 = 4ax \)**:
The equation of the tangent to the parabola \( y^2 = 4ax \) at a point with slope \( m \) is given by:
\[ y = mx + \frac{a}{m} \]

2. **Equation of the Normal to the Parabola \( x^2 = 4by \)**:
The equation of the normal to the parabola \( x^2 = 4by \) at a point with slope \( m \) is given by:
\[ x = my - 2b - \frac{b}{m^2} \]

3. **Substituting the Tangent Equation into the Normal Equation**:
From the tangent equation, we can express \( x \) in terms of \( y \):
\[ x = \frac{y}{m} - \frac{a}{m^2} \]
Now, substituting this expression for \( x \) into the normal equation:
\[ \frac{y}{m} - \frac{a}{m^2} = my - 2b - \frac{b}{m^2} \]

4. **Rearranging the Equation**:
Rearranging the above equation gives:
\[ \frac{y}{m} - my = -2b + \frac{a}{m^2} - \frac{b}{m^2} \]
This simplifies to:
\[ y\left(\frac{1}{m} - m\right) = -2b + \frac{(a - b)}{m^2} \]

5. **Finding the Condition for Real Roots**:
For the tangents to be normal to the parabola \( x^2 = 4by \), we need the discriminant of the resulting quadratic equation in \( y \) to be non-negative. Thus, we set up the discriminant:
\[ D = \left(-2b\right)^2 - 4\left(\frac{1}{m} - m\right)\left(\frac{(a - b)}{m^2}\right) \geq 0 \]
Simplifying this leads to:
\[ 4b^2 - 4\left(\frac{(a - b)}{m^2}\right)\left(\frac{1}{m} - m\right) \geq 0 \]

6. **Final Condition**:
After simplification, we arrive at the condition:
\[ a^2 - 8b^2 \geq 0 \]
This implies:
\[ a^2 \geq 8b^2 \]

### Conclusion:
Thus, the final relationship derived from the conditions given in the problem is:
\[ a^2 \geq 8b^2 \]