Acute angle between two curve `x^(2)+y^(2)=a^(2)sqrt2` and `x^(2)-y^(2)=a^(2)` is

Options

A. `(pi)/(6)`
B. `(pi)/(3)`
C. `(pi)/(4)`
D. none of these

Explanation

To find the acute angle between the curves given by the equations \(x^2 + y^2 = a^2 \sqrt{2}\) and \(x^2 - y^2 = a^2\), we will follow these steps:

### Step 1: Differentiate the equations
We will differentiate both equations with respect to \(x\) to find the slopes of the tangents at the points of intersection.

1. **For the first equation** \(x^2 + y^2 = a^2 \sqrt{2}\):
\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(a^2 \sqrt{2}) \]
This gives:
\[ 2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y} \]
Let \(m_1 = -\frac{x}{y}\).

2. **For the second equation** \(x^2 - y^2 = a^2\):
\[ \frac{d}{dx}(x^2 - y^2) = \frac{d}{dx}(a^2) \]
This gives:
\[ 2x - 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{y} \]
Let \(m_2 = \frac{x}{y}\).

### Step 2: Use the formula for the angle between two curves
The angle \(\theta\) between two curves can be found using the formula:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \(m_1\) and \(m_2\):
\[ \tan \theta = \left| \frac{-\frac{x}{y} - \frac{x}{y}}{1 + \left(-\frac{x}{y}\right) \left(\frac{x}{y}\right)} \right| \]
This simplifies to:
\[ \tan \theta = \left| \frac{-\frac{2x}{y}}{1 - \frac{x^2}{y^2}} \right| = \left| \frac{-2xy}{y^2 - x^2} \right| \]

### Step 3: Determine the points of intersection
To find the points of intersection, we can solve the equations simultaneously. From the second equation:
\[ y^2 = x^2 - a^2 \]
Substituting \(y^2\) into the first equation:
\[ x^2 + (x^2 - a^2) = a^2 \sqrt{2} \]
This simplifies to:
\[ 2x^2 - a^2 = a^2 \sqrt{2} \implies 2x^2 = a^2(\sqrt{2} + 1) \implies x^2 = \frac{a^2(\sqrt{2} + 1)}{2} \]
Thus,
\[ x = a \sqrt{\frac{\sqrt{2} + 1}{2}}, \quad y = \sqrt{x^2 - a^2} = a \sqrt{\frac{\sqrt{2} - 1}{2}} \]

### Step 4: Substitute back to find \(\tan \theta\)
Now substituting \(x\) and \(y\) back into the equation for \(\tan \theta\):
\[ \tan \theta = \left| \frac{-2a \sqrt{\frac{\sqrt{2} + 1}{2}} \cdot a \sqrt{\frac{\sqrt{2} - 1}{2}}}{\left(\frac{\sqrt{2} - 1}{2}\right) - \left(\frac{\sqrt{2} + 1}{2}\right)} \right| \]
This calculation will yield a specific value for \(\tan \theta\).

### Step 5: Find \(\theta\)
Finally, we can find \(\theta\) using:
\[ \theta = \tan^{-1}(\tan \theta) \]
After simplification, we find that:
\[ \theta = \frac{\pi}{4} \]

### Conclusion
The acute angle between the two curves is \(\frac{\pi}{4}\) radians or \(45^\circ\).

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