If `x = a sin 2 theta (1 + cos 2theta ), y=b cos 2 theta (1- cos 2 theta )` then `(dy)/(dx)=`

Options

A. ` ( b tan theta)/( a)`
B. ` ( a tan theta)/(b)`
C. `( b cot theta)/(a)`
D. None

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = a \sin 2\theta (1 + \cos 2\theta)\) and \(y = b \cos 2\theta (1 - \cos 2\theta)\), we will follow these steps:

### Step 1: Differentiate \(y\) with respect to \(\theta\)

We start with the equation for \(y\):
\[ y = b \cos 2\theta (1 - \cos 2\theta) \]
Using the product rule, we differentiate \(y\):
\[ \frac{dy}{d\theta} = b \left( \frac{d}{d\theta}(\cos 2\theta) \cdot (1 - \cos 2\theta) + \cos 2\theta \cdot \frac{d}{d\theta}(1 - \cos 2\theta) \right) \]
Calculating the derivatives:
- \(\frac{d}{d\theta}(\cos 2\theta) = -2\sin 2\theta\)
- \(\frac{d}{d\theta}(1 - \cos 2\theta) = 2\sin 2\theta\)

Substituting these into the equation:
\[ \frac{dy}{d\theta} = b \left( -2\sin 2\theta (1 - \cos 2\theta) + \cos 2\theta \cdot 2\sin 2\theta \right) \]
Simplifying:
\[ \frac{dy}{d\theta} = b \left( -2\sin 2\theta + 2\sin 2\theta \cos 2\theta \right) \]
\[ = 2b \sin 2\theta (\cos 2\theta - 1) \]

### Step 2: Differentiate \(x\) with respect to \(\theta\)

Next, we differentiate \(x\):
\[ x = a \sin 2\theta (1 + \cos 2\theta) \]
Using the product rule again:
\[ \frac{dx}{d\theta} = a \left( \frac{d}{d\theta}(\sin 2\theta) \cdot (1 + \cos 2\theta) + \sin 2\theta \cdot \frac{d}{d\theta}(1 + \cos 2\theta) \right) \]
Calculating the derivatives:
- \(\frac{d}{d\theta}(\sin 2\theta) = 2\cos 2\theta\)
- \(\frac{d}{d\theta}(1 + \cos 2\theta) = -2\sin 2\theta\)

Substituting these into the equation:
\[ \frac{dx}{d\theta} = a \left( 2\cos 2\theta (1 + \cos 2\theta) + \sin 2\theta (-2\sin 2\theta) \right) \]
Simplifying:
\[ \frac{dx}{d\theta} = 2a \cos 2\theta (1 + \cos 2\theta) - 2a \sin^2 2\theta \]
\[ = 2a \left( \cos 2\theta + \cos^2 2\theta - \sin^2 2\theta \right) \]
Using the identity \(\cos^2 2\theta - \sin^2 2\theta = \cos 4\theta\):
\[ \frac{dx}{d\theta} = 2a \left( \cos 2\theta + \cos 4\theta \right) \]

### Step 3: Calculate \(\frac{dy}{dx}\)

Now we can find \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{2b \sin 2\theta (\cos 2\theta - 1)}{2a (\cos 2\theta + \cos 4\theta)} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{b \sin 2\theta (\cos 2\theta - 1)}{a (\cos 2\theta + \cos 4\theta)} \]

### Final Answer
Thus, the final expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{b \sin 2\theta (\cos 2\theta - 1)}{a (\cos 2\theta + \cos 4\theta)} \]