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: Find the slope of the tangent to the ellipse
The given ellipse is:
\[ \frac{x^2}{4} + y^2 = 1 \]
To find the slope of the tangent line to the ellipse, we differentiate the equation implicitly 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:
\[ \frac{x}{2} + 2y \frac{dy}{dx} = 0 \]
Rearranging for \(\frac{dy}{dx}\):
\[ 2y \frac{dy}{dx} = -\frac{x}{2} \]
\[ \frac{dy}{dx} = -\frac{x}{4y} \]
Thus, the slope of the tangent to the ellipse at any point is:
\[ m_1 = -\frac{x}{4y} \]

### Step 2: Use the angle between two curves
We know that the angle \(\theta\) between two curves with slopes \(m_1\) and \(m_2\) is given by:
\[ \tan(\theta) = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| \]
Given that the angle is \(\frac{\pi}{4}\), we have:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
Thus, we set up the equation:
\[ 1 = \left|\frac{-\frac{x}{4y} - m_2}{1 - \frac{x}{4y} m_2}\right| \]

### Step 3: Solve for \(m_2\)
We can drop the absolute value for simplicity and consider:
\[ 1 = \frac{-\frac{x}{4y} - m_2}{1 - \frac{x}{4y} m_2} \]
Cross-multiplying gives:
\[ 1 - \frac{x}{4y} m_2 = -\frac{x}{4y} - m_2 \]
Rearranging terms:
\[ 1 + \frac{x}{4y} = m_2 + \frac{x}{4y} m_2 \]
Factoring out \(m_2\):
\[ 1 + \frac{x}{4y} = m_2\left(1 + \frac{x}{4y}\right) \]
Assuming \(1 + \frac{x}{4y} \neq 0\), we can divide both sides:
\[ m_2 = 1 \]

### Step 4: Express \(m_2\) in terms of \(x\) and \(y\)
We can express \(m_2\) in terms of \(x\) and \(y\):
\[ m_2 = \frac{dy}{dx} = \frac{x + 4y}{x - 4y} \]

### Step 5: Form the differential equation
Thus, we have the differential equation for the family of curves:
\[ \frac{dy}{dx} = \frac{x + 4y}{x - 4y} \]

### Final Answer
The required differential equation is:
\[ \frac{dy}{dx} = \frac{x + 4y}{x - 4y} \]