The number of values of a for which the curves `4x^(2)+a^(2)y^(2)=4a^(2) and y^(2)=16x` are orthogonal is

Answer

To find the number of values of \( a \) for which the curves \( 4x^2 + a^2y^2 = 4a^2 \) and \( y^2 = 16x \) are orthogonal, we need to follow these steps:

### Step 1: Find the slopes of the tangents to both curves

1. **Differentiate the first curve**:
The first curve is given by \( 4x^2 + a^2y^2 = 4a^2 \).
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(4x^2) + \frac{d}{dx}(a^2y^2) = \frac{d}{dx}(4a^2) \]
This gives:
\[ 8x + 2a^2y \frac{dy}{dx} = 0 \]
Rearranging for \( \frac{dy}{dx} \):
\[ 2a^2y \frac{dy}{dx} = -8x \implies \frac{dy}{dx} = -\frac{4x}{a^2y} \]
Thus, the slope \( m_1 \) of the first curve is:
\[ m_1 = -\frac{4x}{a^2y} \]

2. **Differentiate the second curve**:
The second curve is \( y^2 = 16x \).
Differentiating gives:
\[ 2y \frac{dy}{dx} = 16 \implies \frac{dy}{dx} = \frac{8}{y} \]
Thus, the slope \( m_2 \) of the second curve is:
\[ m_2 = \frac{8}{y} \]

### Step 2: Set the condition for orthogonality

For the curves to be orthogonal, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes we found:
\[ \left(-\frac{4x}{a^2y}\right) \cdot \left(\frac{8}{y}\right) = -1 \]
This simplifies to:
\[ -\frac{32x}{a^2y^2} = -1 \implies \frac{32x}{a^2y^2} = 1 \implies 32x = a^2y^2 \]

### Step 3: Substitute \( y^2 \) from the second curve

From the second curve \( y^2 = 16x \), we substitute \( y^2 \) into the equation:
\[ 32x = a^2(16x) \]
This simplifies to:
\[ 32x = 16a^2x \]
Assuming \( x \neq 0 \) (since \( x \) can be any point on the curve), we can divide both sides by \( x \):
\[ 32 = 16a^2 \implies a^2 = 2 \]

### Step 4: Find the values of \( a \)

The equation \( a^2 = 2 \) gives:
\[ a = \sqrt{2} \quad \text{or} \quad a = -\sqrt{2} \]
Thus, there are **2 values** of \( a \).

### Final Answer

The number of values of \( a \) for which the curves are orthogonal is \( \boxed{2} \).