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 two 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 to find the slopes of the tangents.

1. **Differentiate the first curve**:
\[ x^2 + y^2 = a^2 \sqrt{2} \]
Differentiating both sides with respect to \(x\):
\[ 2x + 2y \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = -\frac{x}{y} \]
Let this slope be \(m_1 = -\frac{x}{y}\).

2. **Differentiate the second curve**:
\[ x^2 - y^2 = a^2 \]
Differentiating both sides with respect to \(x\):
\[ 2x - 2y \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = \frac{x}{y} \]
Let this slope be \(m_2 = \frac{x}{y}\).

### Step 2: Use the formula for the angle between two curves.

The formula for the tangent of the angle \(\theta\) between two curves with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]

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:
\[ \tan \theta = \left| \frac{-\frac{2x}{y}}{1 - \frac{x^2}{y^2}} \right| \]

### Step 3: Simplify the expression.

1. The denominator can be rewritten:
\[ 1 - \frac{x^2}{y^2} = \frac{y^2 - x^2}{y^2} \]
Thus,
\[ \tan \theta = \left| \frac{-2xy^2}{y^2 - x^2} \right| \]

### Step 4: Find the values of \(x\) and \(y\) at their meeting points.

To find the meeting points of the curves, we can solve the two equations:
1. From \(x^2 + y^2 = a^2 \sqrt{2}\)
2. From \(x^2 - y^2 = a^2\)

Subtracting the second equation from the first:
\[ (x^2 + y^2) - (x^2 - y^2) = a^2 \sqrt{2} - a^2 \]
This simplifies to:
\[ 2y^2 = a^2 (\sqrt{2} - 1) \]
Thus,
\[ y^2 = \frac{a^2 (\sqrt{2} - 1)}{2} \]

Now substituting \(y^2\) back into one of the original equations to find \(x^2\):
\[ x^2 + \frac{a^2 (\sqrt{2} - 1)}{2} = a^2 \sqrt{2} \]
This gives:
\[ x^2 = a^2 \sqrt{2} - \frac{a^2 (\sqrt{2} - 1)}{2} \]
Simplifying this will yield the value of \(x^2\).

### Step 5: Substitute back to find \(\tan \theta\).

After finding \(x\) and \(y\), substitute back into the expression for \(\tan \theta\):
\[ \tan \theta = \left| \frac{-2xy^2}{y^2 - x^2} \right| \]
This will yield a numerical value.

### Step 6: Calculate the angle \(\theta\).

Finally, calculate \(\theta\) using:
\[ \theta = \tan^{-1}(\tan \theta) \]
If \(\tan \theta = 1\), then \(\theta = 45^\circ\).

### Conclusion

The acute angle between the two curves is \(45^\circ\).

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