The tangents to the curve `x^(2) + y^(2) = 2 ` at points (1,1) and (-1,1) are

Options

A. parallel
B. at right angles
C. neither parallel nor at right angles
D. none of these

Answer

To find the tangents to the curve \( x^2 + y^2 = 2 \) at the points \( (1, 1) \) and \( (-1, 1) \), we will follow these steps:

### Step 1: Differentiate the equation of the curve
We start with the equation of the curve:
\[ x^2 + y^2 = 2 \]
To find the slope of the tangent line, we differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(2) \]
This gives us:
\[ 2x + 2y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to solve for \(\frac{dy}{dx}\):
\[ 2y \frac{dy}{dx} = -2x \]
\[ \frac{dy}{dx} = -\frac{x}{y} \]

### Step 3: Find the slope at the point \( (1, 1) \)
Now we substitute the point \( (1, 1) \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{(1, 1)} = -\frac{1}{1} = -1 \]
So, the slope \( m_1 \) at the point \( (1, 1) \) is \( -1 \).

### Step 4: Find the slope at the point \( (-1, 1) \)
Next, we substitute the point \( (-1, 1) \):
\[ \frac{dy}{dx} \bigg|_{(-1, 1)} = -\frac{-1}{1} = 1 \]
So, the slope \( m_2 \) at the point \( (-1, 1) \) is \( 1 \).

### Step 5: Determine the relationship between the slopes
Now we check the product of the slopes:
\[ m_1 \cdot m_2 = (-1) \cdot 1 = -1 \]
Since the product of the slopes is \( -1 \), this indicates that the tangents at the two points are perpendicular to each other.

### Conclusion
The tangents to the curve \( x^2 + y^2 = 2 \) at the points \( (1, 1) \) and \( (-1, 1) \) are perpendicular.

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