If the curves `ay+x^2=7` and `x^3=y` cut orthogonally at `(1, 1)` then `a=` (A) 1 (B) `-6` (C) 6 (D) `1/6`

Options

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

Explanation

To find the value of \( a \) such that the curves \( ay + x^2 = 7 \) and \( x^3 = y \) cut orthogonally at the point \( (1, 1) \), we will follow these steps:

### Step 1: Differentiate the first curve
The first curve is given by the equation:
\[ ay + x^2 = 7 \]
We will differentiate this equation with respect to \( x \):
\[ \frac{d}{dx}(ay) + \frac{d}{dx}(x^2) = \frac{d}{dx}(7) \]
Using the product rule on \( ay \) and the power rule on \( x^2 \), we get:
\[ a \frac{dy}{dx} + 2x = 0 \]
Rearranging gives:
\[ a \frac{dy}{dx} = -2x \quad \Rightarrow \quad \frac{dy}{dx} = -\frac{2x}{a} \]

### Step 2: Find the slope at the point \( (1, 1) \)
Now, we will evaluate the slope at the point \( (1, 1) \):
\[ \frac{dy}{dx} \bigg|_{(1, 1)} = -\frac{2 \cdot 1}{a} = -\frac{2}{a} \]
Let’s denote this slope as \( m_1 \):
\[ m_1 = -\frac{2}{a} \]

### Step 3: Differentiate the second curve
The second curve is given by:
\[ x^3 = y \]
Differentiating this with respect to \( x \) gives:
\[ 3x^2 = \frac{dy}{dx} \]
Let’s denote this slope as \( m_2 \):
\[ m_2 = 3x^2 \]

### Step 4: Find the slope at the point \( (1, 1) \)
Now, we will evaluate the slope at the point \( (1, 1) \):
\[ m_2 \bigg|_{(1, 1)} = 3 \cdot 1^2 = 3 \]

### Step 5: Set up the orthogonality condition
The curves cut 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:
\[ -\frac{6}{a} = -1 \]
Removing the negative signs gives:
\[ \frac{6}{a} = 1 \]

### Step 6: Solve for \( a \)
Cross-multiplying gives:
\[ 6 = a \]
Thus, the value of \( a \) is:
\[ \boxed{6} \]