Let a curve passes through (3, 2) and satisfied the differential equation (x-1) dx + 4(y -2) dy = 0

Options

A. It represents equation of a circle
B. It represents equation of an ellipse
C. Area euclosed by the curve is `2pi`.
D. Line y = 1 is a tengent to the curve

Answer

To solve the given problem, we need to find the equation of the curve that passes through the point (3, 2) and satisfies the differential equation \((x-1)dx + 4(y-2)dy = 0\).

### Step-by-Step Solution:

1. **Rearranging the Differential Equation:**
We start with the given differential equation:
\[ (x - 1)dx + 4(y - 2)dy = 0 \]
Rearranging gives:
\[ (x - 1)dx = -4(y - 2)dy \]

2. **Separating Variables:**
We can separate the variables:
\[ \frac{dx}{4(y - 2)} = -\frac{dy}{(x - 1)} \]

3. **Integrating Both Sides:**
Now we integrate both sides:
\[ \int \frac{dx}{(x - 1)} = -4 \int \frac{dy}{(y - 2)} \]
The left side integrates to:
\[ \ln |x - 1| \]
The right side integrates to:
\[ -4 \ln |y - 2| \]
Thus, we have:
\[ \ln |x - 1| = -4 \ln |y - 2| + C \]

4. **Exponentiating Both Sides:**
To eliminate the logarithm, we exponentiate both sides:
\[ |x - 1| = e^C |y - 2|^{-4} \]
Let \(k = e^C\), then we can write:
\[ |x - 1| = \frac{k}{(y - 2)^4} \]

5. **Finding the Constant \(k\):**
Since the curve passes through the point (3, 2), we substitute \(x = 3\) and \(y = 2\):
\[ |3 - 1| = \frac{k}{(2 - 2)^4} \]
This leads to an undefined expression since the denominator becomes zero. We need to adjust our approach to find the specific form of the equation.

6. **Revisiting the Integration:**
From the integration, we can write:
\[ (x - 1)^2 = -4(y - 2)^2 + C \]
Rearranging gives:
\[ \frac{(x - 1)^2}{4} + \frac{(y - 2)^2}{1} = C \]

7. **Substituting the Point (3, 2):**
Substituting \(x = 3\) and \(y = 2\) into the equation:
\[ \frac{(3 - 1)^2}{4} + \frac{(2 - 2)^2}{1} = C \]
This simplifies to:
\[ \frac{4}{4} + 0 = C \implies C = 1 \]

8. **Final Equation of the Curve:**
Thus, the equation of the curve is:
\[ \frac{(x - 1)^2}{4} + (y - 2)^2 = 1 \]
This represents an ellipse centered at (1, 2) with semi-major axis 2 and semi-minor axis 1.