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 solve the problem of finding the number of values of \( a \) for which the curves \( 4x^2 + a^2y^2 = 4a^2 \) and \( y^2 = 16x \) are orthogonal, we will follow these steps:

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

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

2. **Differentiate the second curve**:
The second curve is given by:
\[ y^2 = 16x \]
Differentiating both sides with respect to \( x \):
\[ 2y \frac{dy}{dx} = 16 \]
Rearranging gives:
\[ \frac{dy}{dx} = \frac{16}{2y} = \frac{8}{y} \]
Let \( m_2 = \frac{8}{y} \).

### Step 2: Set up the condition for orthogonality.

The curves are orthogonal if the product of their slopes at the point of intersection is equal to -1:
\[ m_1 \cdot m_2 = -1 \]
Substituting the expressions for \( m_1 \) and \( m_2 \):
\[ \left(-\frac{4x}{a^2y}\right) \cdot \left(\frac{8}{y}\right) = -1 \]
This simplifies to:
\[ -\frac{32x}{a^2y^2} = -1 \implies \frac{32x}{a^2y^2} = 1 \]
Thus, we have:
\[ 32x = a^2y^2 \]

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

From the second curve \( y^2 = 16x \), we can substitute \( y^2 \) into the equation:
\[ 32x = a^2(16x) \]
Dividing both sides by \( x \) (assuming \( x \neq 0 \)):
\[ 32 = 16a^2 \implies a^2 = 2 \]

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

Taking the square root of both sides gives:
\[ a = \pm \sqrt{2} \]

### Conclusion

Thus, the number of values of \( a \) for which the curves are orthogonal is **2** (i.e., \( a = \sqrt{2} \) and \( a = -\sqrt{2} \)).