The angle of intersection of the curves `xy = a ^(2) and x ^(2) - y ^(2) = 2 a ^(2)` is :

Options

A. `0^(@)`
B. `45 ^(@)`
C. `90^(@)`
D. `30^(@)`

Answer

To find the angle of intersection of the curves given by the equations \( xy = a^2 \) and \( x^2 - y^2 = 2a^2 \), we will follow these steps:

### Step 1: Differentiate the first curve
The first curve is given by:
\[ xy = a^2 \]

To find the slope of the tangent to this curve, we differentiate implicitly with respect to \( x \):
\[ \frac{d}{dx}(xy) = \frac{d}{dx}(a^2) \]
Using the product rule, we have:
\[ y + x \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ \frac{dy}{dx} = -\frac{y}{x} \]
Let this slope be \( M_1 \):
\[ M_1 = -\frac{y}{x} \]

### Step 2: Differentiate the second curve
The second curve is given by:
\[ x^2 - y^2 = 2a^2 \]

Differentiating implicitly with respect to \( x \):
\[ \frac{d}{dx}(x^2) - \frac{d}{dx}(y^2) = \frac{d}{dx}(2a^2) \]
This gives:
\[ 2x - 2y \frac{dy}{dx} = 0 \]
Rearranging gives:
\[ 2y \frac{dy}{dx} = 2x \quad \Rightarrow \quad \frac{dy}{dx} = \frac{x}{y} \]
Let this slope be \( M_2 \):
\[ M_2 = \frac{x}{y} \]

### Step 3: Find the product of the slopes
Now we find the product of the slopes \( M_1 \) and \( M_2 \):
\[ M_1 \cdot M_2 = \left(-\frac{y}{x}\right) \cdot \left(\frac{x}{y}\right) \]
This simplifies to:
\[ M_1 \cdot M_2 = -1 \]

### Step 4: Determine the angle of intersection
The condition for two curves to intersect at right angles (90 degrees) is:
\[ M_1 \cdot M_2 = -1 \]
Since we have found that \( M_1 \cdot M_2 = -1 \), it follows that the angle \( \theta \) between the two curves is:
\[ \theta = 90^\circ \]

### Final Answer
Thus, the angle of intersection of the curves \( xy = a^2 \) and \( x^2 - y^2 = 2a^2 \) is:
\[ \theta = 90^\circ \]