The differential equation of the family of curves whose tangent at any point makes an angle of `(pi)/(4)` with the ellipse `(x^(2))/(4)+y^(2)=1` is

Options

A. `(dy)/(dx)=(x+y)/(x-y)`
B. `(dy)/(dx)=(x+4y)/(x-4y)`
C. `(dy)/(dx)=(x)/(4y)`
D. `(dy)/(dx)=(4y)/(x)`

Answer

To find the differential equation of the family of curves whose tangent at any point makes an angle of \(\frac{\pi}{4}\) with the ellipse \(\frac{x^2}{4} + y^2 = 1\), we can follow these steps:

### Step 1: Differentiate the equation of the ellipse
We start with the equation of the ellipse:
\[ \frac{x^2}{4} + y^2 = 1 \]
Differentiating both sides with respect to \(x\):
\[ \frac{d}{dx}\left(\frac{x^2}{4}\right) + \frac{d}{dx}(y^2) = 0 \]
This gives:
\[ \frac{2x}{4} + 2y \frac{dy}{dx} = 0 \]
Simplifying, we have:
\[ \frac{x}{2} + 2y \frac{dy}{dx} = 0 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Rearranging the equation to isolate \(\frac{dy}{dx}\):
\[ 2y \frac{dy}{dx} = -\frac{x}{2} \]
\[ \frac{dy}{dx} = -\frac{x}{4y} \]

### Step 3: Use the angle condition
Let \(m_1\) be the slope of the tangent to the family of curves, and \(m_2\) be the slope of the tangent to the ellipse. We know that the angle between the two tangents is \(\frac{\pi}{4}\). The formula for the tangent of the angle between two lines is:
\[ \tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]
Setting \(\theta = \frac{\pi}{4}\), we have:
\[ 1 = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]
This gives us two cases to consider:
1. \(m_1 - m_2 = 1 + m_1 m_2\)
2. \(m_1 - m_2 = - (1 + m_1 m_2)\)

### Step 4: Substitute \(m_2\)
From Step 2, we have \(m_2 = -\frac{x}{4y}\). We can substitute this into both cases.

#### Case 1:
\[ m_1 + \frac{x}{4y} = 1 + m_1 \left(-\frac{x}{4y}\right) \]
Rearranging gives:
\[ m_1 + \frac{x}{4y} = 1 - \frac{m_1 x}{4y} \]
Multiplying through by \(4y\) to eliminate the denominator:
\[ 4y m_1 + x = 4y - m_1 x \]
Rearranging yields:
\[ (4y + x)m_1 = 4y - x \]
Thus:
\[ m_1 = \frac{4y - x}{4y + x} \]

#### Case 2:
\[ m_1 + \frac{x}{4y} = -1 - m_1 \left(-\frac{x}{4y}\right) \]
This leads to a similar equation after simplification.

### Step 5: Express \(m_1\) as \(\frac{dy}{dx}\)
Since \(m_1 = \frac{dy}{dx}\), we have:
\[ \frac{dy}{dx} = \frac{4y - x}{4y + x} \]

### Step 6: Form the differential equation
This gives us the differential equation:
\[ \frac{dy}{dx} = \frac{4y - x}{4y + x} \]

### Final Step: Rearranging
Rearranging the equation leads to:
\[ (4y + x) \frac{dy}{dx} = 4y - x \]
This is the required differential equation.