If the curves ay `+ x^(2) = 7 and y = x^(3)` cut orthogonally at (1,1) , then the value of a is

Options

A. 1
B. 0
C. 6
D. `-6`

Answer

To solve the problem, we need to find the value of \( a \) such that the curves \( ay + x^2 = 7 \) and \( y = x^3 \) intersect orthogonally at the point \( (1, 1) \).

### Step-by-step Solution:

1. **Identify the equations of the curves**:
- The first curve is given by \( ay + x^2 = 7 \).
- The second curve is given by \( y = x^3 \).

2. **Differentiate the first curve**:
- Rearranging the first equation gives \( y = \frac{7 - x^2}{a} \).
- Now, differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{7 - x^2}{a}\right) = \frac{-2x}{a} \]
- Let this slope be \( m_1 = \frac{-2x}{a} \).

3. **Differentiate the second curve**:
- The second curve \( y = x^3 \) can be differentiated directly:
\[ \frac{dy}{dx} = 3x^2 \]
- Let this slope be \( m_2 = 3x^2 \).

4. **Evaluate the slopes at the point (1, 1)**:
- For \( m_1 \) at \( (1, 1) \):
\[ m_1 = \frac{-2(1)}{a} = \frac{-2}{a} \]
- For \( m_2 \) at \( (1, 1) \):
\[ m_2 = 3(1^2) = 3 \]

5. **Use the condition for orthogonality**:
- The curves intersect orthogonally if the product of their slopes is \( -1 \):
\[ m_1 \cdot m_2 = -1 \]
- Substituting the values of \( m_1 \) and \( m_2 \):
\[ \left(\frac{-2}{a}\right) \cdot 3 = -1 \]
- This simplifies to:
\[ \frac{-6}{a} = -1 \]

6. **Solve for \( a \)**:
- Multiply both sides by \( a \):
\[ -6 = -a \]
- Therefore:
\[ a = 6 \]

### Final Answer:
The value of \( a \) is \( 6 \).