The angle between the curves `x^(2)+4y^(2)=32 and x^(2)-y^(2)=12,` is

Options

A. `(pi)/(3)`
B. `(pi)/(4)`
C. `(pi)/(6)`
D. `(pi)/(2)`

Answer

To find the angle between the curves given by the equations \(x^2 + 4y^2 = 32\) and \(x^2 - y^2 = 12\), we will follow these steps:

### Step 1: Find the Points of Intersection
We need to find the points where the two curves intersect.

1. From the first equation, we can express \(x^2\) in terms of \(y\):
\[ x^2 = 32 - 4y^2 \]

2. Substitute this expression for \(x^2\) into the second equation:
\[ (32 - 4y^2) - y^2 = 12 \]
Simplifying this gives:
\[ 32 - 5y^2 = 12 \]
\[ 5y^2 = 20 \quad \Rightarrow \quad y^2 = 4 \quad \Rightarrow \quad y = \pm 2 \]

3. Now substitute \(y = 2\) and \(y = -2\) back into the equation for \(x^2\):
- For \(y = 2\):
\[ x^2 = 32 - 4(2^2) = 32 - 16 = 16 \quad \Rightarrow \quad x = \pm 4 \]
- For \(y = -2\):
\[ x^2 = 32 - 4(-2)^2 = 32 - 16 = 16 \quad \Rightarrow \quad x = \pm 4 \]

Thus, the points of intersection are \((4, 2)\), \((-4, 2)\), \((4, -2)\), and \((-4, -2)\).

### Step 2: Find the Slopes of the Tangents at the Points of Intersection
Next, we will find the slopes of the tangents to both curves at one of the intersection points, say \((4, 2)\).

1. For the first curve \(x^2 + 4y^2 = 32\):
- Differentiate implicitly:
\[ 2x + 8yy' = 0 \quad \Rightarrow \quad y' = -\frac{x}{4y} \]
- At the point \((4, 2)\):
\[ y' = -\frac{4}{4 \cdot 2} = -\frac{4}{8} = -\frac{1}{2} \]
Let \(m_1 = -\frac{1}{2}\).

2. For the second curve \(x^2 - y^2 = 12\):
- Differentiate implicitly:
\[ 2x - 2yy' = 0 \quad \Rightarrow \quad y' = \frac{x}{y} \]
- At the point \((4, 2)\):
\[ y' = \frac{4}{2} = 2 \]
Let \(m_2 = 2\).

### Step 3: Calculate the Angle Between the Tangents
The angle \(\theta\) between the two tangents can be found using the formula:
\[ \tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| \]
Substituting \(m_1\) and \(m_2\):
\[ \tan(\theta) = \left| \frac{-\frac{1}{2} - 2}{1 + (-\frac{1}{2})(2)} \right| = \left| \frac{-\frac{1}{2} - \frac{4}{2}}{1 - 1} \right| = \left| \frac{-\frac{5}{2}}{0} \right| \]
Since the denominator is zero, this indicates that the tangents are perpendicular.

Thus, the angle \(\theta\) is:
\[ \theta = \frac{\pi}{2} \text{ radians} \quad \text{or} \quad 90^\circ \]

### Final Answer
The angle between the curves is \(\frac{\pi}{2}\).

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