Find the condition for the curve `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 and xy =c^(2)` to interest orthogonally.

Answer

To find the condition for the curves \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and \(xy = c^2\) to intersect orthogonally, we will follow these steps:

### Step 1: Find the slope of the first curve
We start with the first curve:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}\left(\frac{x^2}{a^2}\right) - \frac{d}{dx}\left(\frac{y^2}{b^2}\right) = 0 \]
Using the chain rule, we have:
\[ \frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{2y}{b^2} \frac{dy}{dx} = \frac{2x}{a^2} \]
Thus:
\[ \frac{dy}{dx} = \frac{x}{y} \cdot \frac{b^2}{a^2} \]
Let \(m_1 = \frac{x}{y} \cdot \frac{b^2}{a^2}\) be the slope of the first curve.

### Step 2: Find the slope of the second curve
Now consider the second curve:
\[ xy = c^2 \]
Differentiating both sides with respect to \(x\):
\[ y + x \frac{dy}{dx} = 0 \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{y}{x} \]
Let \(m_2 = -\frac{y}{x}\) be the slope of the second curve.

### Step 3: Set up the orthogonality condition
For the curves to intersect orthogonally, the product of their slopes must equal \(-1\):
\[ m_1 \cdot m_2 = -1 \]
Substituting the expressions for \(m_1\) and \(m_2\):
\[ \left(\frac{x}{y} \cdot \frac{b^2}{a^2}\right) \left(-\frac{y}{x}\right) = -1 \]
This simplifies to:
\[ -\frac{b^2}{a^2} = -1 \]
Thus:
\[ \frac{b^2}{a^2} = 1 \]

### Step 4: Final condition
From \(\frac{b^2}{a^2} = 1\), we conclude:
\[ b^2 = a^2 \]
This implies that:
\[ a = b \quad \text{or} \quad a = -b \]
However, since \(a\) and \(b\) represent lengths, we take:
\[ a = b \]

### Conclusion
The condition for the curves \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and \(xy = c^2\) to intersect orthogonally is:
\[ a = b \]