The differential equation of the family of ellipses having major and minor axes respectively along the x and y-axes and minor axis is equal to half of the major axis, is

Options

A. `xy'-4y=0`
B. `4xy'+y=0`
C. `4yy'+x=0`
D. `yy'+4x=0`

Answer

To find the differential equation of the family of ellipses where the major and minor axes are along the x and y axes respectively, and the minor axis is equal to half of the major axis, we can follow these steps:

### Step 1: Write the standard equation of the ellipse
The standard form of an ellipse with the major axis along the x-axis and the minor axis along the y-axis is given by:
\[ \frac{y^2}{b^2} + \frac{x^2}{a^2} = 1 \]
where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

### Step 2: Relate the axes
According to the problem, the minor axis is equal to half of the major axis. This implies:
\[ b = \frac{1}{2} a \]

### Step 3: Substitute \( b \) in the ellipse equation
Substituting \( b = \frac{1}{2} a \) into the ellipse equation, we get:
\[ \frac{y^2}{\left(\frac{1}{2} a\right)^2} + \frac{x^2}{a^2} = 1 \]
This simplifies to:
\[ \frac{y^2}{\frac{1}{4} a^2} + \frac{x^2}{a^2} = 1 \]
Multiplying through by \( 4a^2 \) to eliminate the denominators gives:
\[ 4y^2 + 4x^2 = 4a^2 \]
or
\[ 4y^2 + 4x^2 = 4a^2 \]

### Step 4: Differentiate the equation
Now, we differentiate both sides of the equation \( 4y^2 + 4x^2 = 4a^2 \) with respect to \( x \):
\[ \frac{d}{dx}(4y^2) + \frac{d}{dx}(4x^2) = \frac{d}{dx}(4a^2) \]
Using the chain rule, we have:
\[ 4 \cdot 2y \frac{dy}{dx} + 4 \cdot 2x = 0 \]
This simplifies to:
\[ 8y \frac{dy}{dx} + 8x = 0 \]

### Step 5: Simplify the equation
Dividing the entire equation by 8 gives:
\[ y \frac{dy}{dx} + x = 0 \]
or rearranging gives:
\[ 4y \frac{dy}{dx} + x = 0 \]

### Final Result
Thus, the differential equation of the family of ellipses is:
\[ 4y \frac{dy}{dx} + x = 0 \]