The parabolas `y^2=4ax` and `x^2=4by` intersect orthogonally at point `P(x_1,y_1)` where `x_1,y_1 != 0` provided (A) `b=a^2` (B) `b=a^3` (C) `b^3=a^2` (D) none of these

Options

A. `b = a^(2)`
B. `b = a^(3)`
C. `b^(3) = a^(2)`
D. None of these

Explanation

To solve the problem, we need to find the relationship between the parameters \( a \) and \( b \) for the parabolas \( y^2 = 4ax \) and \( x^2 = 4by \) that intersect orthogonally at a point \( P(x_1, y_1) \) where both \( x_1 \) and \( y_1 \) are non-zero.

### Step-by-Step Solution:

1. **Find the slopes of the tangents to the parabolas:**
- For the first parabola \( y^2 = 4ax \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4ax) \implies 2y \frac{dy}{dx} = 4a \implies \frac{dy}{dx} = \frac{2a}{y} \]
Let \( m_1 = \frac{2a}{y_1} \) (slope of the tangent to the first parabola at point \( P \)).

- For the second parabola \( x^2 = 4by \):
\[ \frac{d}{dx}(x^2) = \frac{d}{dx}(4by) \implies 2x = 4b \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{x}{2b} \]
Let \( m_2 = \frac{x_1}{2b} \) (slope of the tangent to the second parabola at point \( P \)).

2. **Condition for orthogonality:**
- The parabolas intersect orthogonally if the product of their slopes is \(-1\):
\[ m_1 \cdot m_2 = -1 \implies \left(\frac{2a}{y_1}\right) \cdot \left(\frac{x_1}{2b}\right) = -1 \]
- Simplifying gives:
\[ \frac{ax_1}{by_1} = -1 \implies ax_1 = -by_1 \implies x_1 = -\frac{b}{a}y_1 \]

3. **Substituting into the first parabola's equation:**
- Substitute \( x_1 \) into \( y^2 = 4ax \):
\[ y_1^2 = 4a\left(-\frac{b}{a}y_1\right) \implies y_1^2 + 4by_1 = 0 \]
- Factoring gives:
\[ y_1(y_1 + 4b) = 0 \]
- Since \( y_1 \neq 0 \), we have:
\[ y_1 + 4b = 0 \implies y_1 = -4b \]

4. **Finding \( x_1 \):**
- Substitute \( y_1 = -4b \) into \( x_1 = -\frac{b}{a}y_1 \):
\[ x_1 = -\frac{b}{a}(-4b) = \frac{4b^2}{a} \]

5. **Substituting into the second parabola's equation:**
- Substitute \( x_1 \) and \( y_1 \) into \( x^2 = 4by \):
\[ \left(\frac{4b^2}{a}\right)^2 = 4b(-4b) \implies \frac{16b^4}{a^2} = -16b^2 \]
- Rearranging gives:
\[ 16b^4 + 16a^2b^2 = 0 \implies b^2( b^2 + a^2) = 0 \]
- Since \( b^2 \neq 0 \) (as \( b \) cannot be zero), we have:
\[ b^2 + a^2 = 0 \]
- This implies \( b^2 = -a^2 \), which is not possible for real \( a \) and \( b \).

6. **Conclusion:**
- Since \( b^2 = -a^2 \) is not valid for real numbers, the only conclusion is that the relationship does not fit any of the provided options.

### Final Answer:
The correct option is (D) none of these.