Find the angle of intersection of the curves `x y=a^2a n dx^2+y^2=2a^2`

Answer

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

### Step 1: Identify the curves
The first curve is given by:
\[ xy = a^2 \]
The second curve is given by:
\[ x^2 + y^2 = 2a^2 \]

### Step 2: Find the points of intersection
To find the points of intersection, we can substitute \(y\) from the first equation into the second equation. From \(xy = a^2\), we have:
\[ y = \frac{a^2}{x} \]
Substituting this into the second equation:
\[ x^2 + \left(\frac{a^2}{x}\right)^2 = 2a^2 \]
This simplifies to:
\[ x^2 + \frac{a^4}{x^2} = 2a^2 \]
Multiplying through by \(x^2\) to eliminate the fraction:
\[ x^4 - 2a^2x^2 + a^4 = 0 \]
Letting \(u = x^2\), we have:
\[ u^2 - 2a^2u + a^4 = 0 \]
Using the quadratic formula:
\[ u = \frac{2a^2 \pm \sqrt{(2a^2)^2 - 4 \cdot 1 \cdot a^4}}{2 \cdot 1} = \frac{2a^2 \pm \sqrt{4a^4 - 4a^4}}{2} = \frac{2a^2}{2} = a^2 \]
Thus, \(x^2 = a^2\) implies \(x = a\) or \(x = -a\).

Substituting \(x = a\) into \(y = \frac{a^2}{x}\):
\[ y = \frac{a^2}{a} = a \quad \text{(Point: (a, a))} \]
Substituting \(x = -a\):
\[ y = \frac{a^2}{-a} = -a \quad \text{(Point: (-a, -a))} \]

### Step 3: Find the derivatives (slopes of tangents)
For the first curve \(xy = a^2\):
\[ \frac{dy}{dx} = -\frac{y}{x} \]
At the point \((a, a)\):
\[ \frac{dy}{dx} = -\frac{a}{a} = -1 \]
At the point \((-a, -a)\):
\[ \frac{dy}{dx} = -\frac{-a}{-a} = -1 \]

For the second curve \(x^2 + y^2 = 2a^2\):
\[ 2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y} \]
At the point \((a, a)\):
\[ \frac{dy}{dx} = -\frac{a}{a} = -1 \]
At the point \((-a, -a)\):
\[ \frac{dy}{dx} = -\frac{-a}{-a} = -1 \]

### Step 4: Calculate the angles of the tangents
Let \(\theta_1\) and \(\theta_2\) be the angles made by the tangents with the x-axis. Since both slopes are \(-1\):
\[ \tan(\theta_1) = -1 \implies \theta_1 = \tan^{-1}(-1) = \frac{3\pi}{4} \text{ (or } 135^\circ\text{)} \]
\[ \tan(\theta_2) = -1 \implies \theta_2 = \tan^{-1}(-1) = \frac{3\pi}{4} \text{ (or } 135^\circ\text{)} \]

### Step 5: Find the angle of intersection
The angle \(\theta\) between the two curves is given by:
\[ \theta = |\theta_1 - \theta_2| = |135^\circ - 135^\circ| = 0^\circ \]
Thus, the angle of intersection at the point \((a, a)\) or \((-a, -a)\) is:
\[ \theta = 0^\circ \]

### Final Answer
The angle of intersection of the curves \(xy = a^2\) and \(x^2 + y^2 = 2a^2\) is \(0^\circ\).