Equation to the curve through (2, 1) whose slope at the point (x, y) is `(x^(2)+y^(2))//2xy` is

Options

A. `2(x^(2)-y^(2))=3x`
B. `2(y^(2)-x^(2))=6y`
C. `x(x^(2)-y^(2))=6`
D. None of these

Answer

To find the equation of the curve passing through the point (2, 1) with the given slope at any point (x, y) as \(\frac{x^2 + y^2}{2xy}\), we can follow these steps:

### Step 1: Set up the differential equation
The slope of the curve at any point (x, y) is given by:
\[ \frac{dy}{dx} = \frac{x^2 + y^2}{2xy} \]

### Step 2: Substitute \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we have:
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Substituting this into the differential equation gives:
\[ v + x\frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2 + v^2x^2}{2vx^2} = \frac{1 + v^2}{2v} \]

### Step 3: Rearranging the equation
Rearranging the equation, we have:
\[ v + x\frac{dv}{dx} = \frac{1 + v^2}{2v} \]
This simplifies to:
\[ x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v \]
\[ x\frac{dv}{dx} = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v} \]

### Step 4: Separate variables
Now, we separate the variables:
\[ \frac{2v}{1 - v^2} dv = \frac{dx}{x} \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \int \frac{2v}{1 - v^2} dv = \int \frac{dx}{x} \]
The left side can be integrated using the substitution \(u = 1 - v^2\), giving:
\[ -\ln|1 - v^2| = \ln|x| + C \]

### Step 6: Exponentiate to eliminate logarithms
Exponentiating both sides results in:
\[ |1 - v^2| = \frac{K}{x} \]
where \(K = e^{-C}\).

### Step 7: Substitute back for \(v\)
Recall that \(v = \frac{y}{x}\), so we substitute back:
\[ |1 - \left(\frac{y}{x}\right)^2| = \frac{K}{x} \]
This leads to:
\[ 1 - \frac{y^2}{x^2} = \frac{K}{x} \quad \text{or} \quad \frac{y^2}{x^2} = 1 - \frac{K}{x} \]

### Step 8: Solve for \(K\) using the point (2, 1)
Since the curve passes through (2, 1):
\[ \frac{1^2}{2^2} = 1 - \frac{K}{2} \]
This simplifies to:
\[ \frac{1}{4} = 1 - \frac{K}{2} \]
\[ \frac{K}{2} = 1 - \frac{1}{4} = \frac{3}{4} \]
Thus, \(K = \frac{3}{2}\).

### Step 9: Substitute \(K\) back into the equation
Substituting \(K\) back gives:
\[ \frac{y^2}{x^2} = 1 - \frac{3/2}{x} \]
Multiplying through by \(x^2\):
\[ y^2 = x^2 - \frac{3}{2}x \]

### Final Step: Rearranging to standard form
Rearranging gives:
\[ 2y^2 = 2x^2 - 3x \]
or
\[ 2x^2 - 2y^2 - 3x = 0 \]

### Summary
The equation of the curve is:
\[ 2x^2 - 2y^2 - 3x = 0 \]