The equation of the family of curves which intersect the hyperbola xy=2 orthogonally is

Options

A. `y= (x^3)/(6) +C`
B. `y=(x^2)/(4)+C`
C. `Y=-(x^3)/(6)+C`
D. `y=-(x^2)/( 4)+C`

Answer

To find the equation of the family of curves that intersect the hyperbola \( xy = 2 \) orthogonally, we can follow these steps:

### Step 1: Understand the condition for orthogonality
Two curves intersect orthogonally if the product of their slopes at the point of intersection is \(-1\). If we denote the family of curves as \( f(x, y) = 0 \) and the hyperbola as \( g(x, y) = xy - 2 = 0 \), then at the point of intersection, we have:
\[ f'(x) \cdot g'(x) = -1 \]

### Step 2: Find the slope of the hyperbola
The hyperbola is given by \( xy = 2 \). We can express \( y \) in terms of \( x \):
\[ y = \frac{2}{x} \]
Now, we differentiate this with respect to \( x \):
\[ \frac{dy}{dx} = -\frac{2}{x^2} \]

### Step 3: Set up the orthogonality condition
Let the slope of the family of curves at the point of intersection be \( \frac{dy}{dx} = m \). The orthogonality condition gives us:
\[ m \cdot \left(-\frac{2}{x^2}\right) = -1 \]
This simplifies to:
\[ m = \frac{x^2}{2} \]

### Step 4: Integrate to find the family of curves
Since \( m = \frac{dy}{dx} \), we have:
\[ \frac{dy}{dx} = \frac{x^2}{2} \]
Now, we integrate both sides with respect to \( x \):
\[ y = \int \frac{x^2}{2} \, dx = \frac{1}{2} \cdot \frac{x^3}{3} + C = \frac{x^3}{6} + C \]

### Step 5: Write the final equation
Thus, the equation of the family of curves that intersect the hyperbola \( xy = 2 \) orthogonally is:
\[ y = \frac{x^3}{6} + C \]