The slope of tangent at (x,y) to a curve passing through (2, 1) is `(x^(2)+y^(2))/(2xy)`, then the equation of the curve is

Options

A. `x(x^(2) + y^(2)) = 10`
B. `x(x^(2) - y^(2)) = 6`
C. `2(x^(2) - y^(2)) = 6y`
D. `2(x^(2) - y^(2)) = 3x`

Answer

To solve the problem, we start with the given slope of the tangent to the curve at the point (x, y):

\[ \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \]

### Step 1: Rewrite the equation
We can rearrange the equation to separate variables. Multiply both sides by \(2xy\):

\[ 2xy \frac{dy}{dx} = x^2 + y^2 \]

### Step 2: Rearranging the equation
Next, we can rearrange the equation to isolate the terms involving \(y\) and \(x\):

\[ 2xy \, dy = (x^2 + y^2) \, dx \]

### Step 3: Divide by \(xy\)
Now, we can divide both sides by \(xy\):

\[ 2 \, dy = \left(\frac{x^2}{xy} + \frac{y^2}{xy}\right) \, dx \]

This simplifies to:

\[ 2 \, dy = \left(\frac{x}{y} + \frac{y}{x}\right) \, dx \]

### Step 4: Further simplification
Now, we can rewrite the equation:

\[ 2 \, dy = \left(\frac{x}{y} + \frac{y}{x}\right) \, dx \]

### Step 5: Integrating both sides
To solve this, we can integrate both sides. The left side integrates to \(2y\), and the right side can be integrated as follows:

\[ \int 2 \, dy = \int \left(\frac{x}{y} + \frac{y}{x}\right) \, dx \]

The right side can be split into two integrals:

\[ 2y = \int \frac{x}{y} \, dx + \int \frac{y}{x} \, dx \]

### Step 6: Solving the integrals
The integral of \(\frac{x}{y}\) with respect to \(x\) is \(\frac{x^2}{2y}\) and the integral of \(\frac{y}{x}\) with respect to \(x\) is \(y \ln |x|\):

\[ 2y = \frac{x^2}{2y} + y \ln |x| + C \]

### Step 7: Rearranging the equation
Now, we can rearrange this equation to express \(y\) in terms of \(x\):

\[ 2y^2 = x^2 + 2y \ln |x| + 2C \]

### Step 8: Final form
This can be rearranged to form a quadratic equation in \(y\):

\[ 2y^2 - 2y \ln |x| - x^2 - 2C = 0 \]

### Step 9: Using the point (2, 1)
We know the curve passes through the point (2, 1). We can substitute \(x = 2\) and \(y = 1\) to find the constant \(C\):

\[ 2(1)^2 - 2(1) \ln |2| - (2)^2 - 2C = 0 \]

This simplifies to:

\[ 2 - 2 \ln 2 - 4 - 2C = 0 \]

So,

\[ -2 - 2 \ln 2 - 2C = 0 \implies C = -1 - \ln 2 \]

### Conclusion
Thus, the equation of the curve is:

\[ 2y^2 - 2y \ln |x| - x^2 + 2(1 + \ln 2) = 0 \]