If $x=a(1+cos~2 heta)sin~2 heta; y=b(1-cos~2 heta)cos~2 heta$ then $\frac{dy}{dx}=$

Q: If $x=a(1+cos~2\theta)sin~2\theta; y=b(1-cos~2\theta)cos~2\theta$ then $\frac{dy}{dx}=$

If $x=a(1+cos~2\theta)sin~2\theta; y=b(1-cos~2\theta)cos~2\theta$ then $\frac{dy}{dx}=$

Question

If $x=a(1+cos~2	heta)sin~2	heta; y=b(1-cos~2	heta)cos~2	heta$ then $\frac{dy}{dx}=$

Accepted Answer

To determine $\frac{dy}{dx}$ for the given parametric equations $x = a(1 + \cos 2\theta) \sin 2\theta$ and $y = b(1 - \cos 2\theta) \cos 2\theta$, we need to find the derivatives $\frac{dx}{d\theta}$ and $\frac{dy}{d\theta}$ and then use the chain rule to find $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$.

**Step 1: Differentiate $x$ with respect to $\theta$:**

$$ x = a(1 + \cos 2\theta) \sin 2\theta $$

Using the product rule:

$$ \frac{dx}{d\theta} = a \left[ \frac{d}{d\theta}(1 + \cos 2\theta) \cdot \sin 2\theta + (1 + \cos 2\theta) \cdot \frac{d}{d\theta}(\sin 2\theta) \right] $$

$$ = a \left[ (-2\sin 2\theta) \cdot \sin 2\theta + (1 + \cos 2\theta) \cdot 2\cos 2\theta \right] $$

$$ = a \left[ -2\sin^2 2\theta + 2\cos 2\theta + 2\cos^2 2\theta \right] $$

$$ = a \left[ 2\cos 2\theta - 2\sin^2 2\theta + 2\cos^2 2\theta \right] $$

Using the identity $\sin^2 2\theta + \cos^2 2\theta = 1$:

$$ = a \left[ 2\cos 2\theta + 2(\cos^2 2\theta - \sin^2 2\theta) \right] $$

**Step 2: Differentiate $y$ with respect to $\theta$:**

$$ y = b(1 - \cos 2\theta) \cos 2\theta $$

Using the product rule:

$$ \frac{dy}{d\theta} = b \left[ \frac{d}{d\theta}(1 - \cos 2\theta) \cdot \cos 2\theta + (1 - \cos 2\theta) \cdot \frac{d}{d\theta}(\cos 2\theta) \right] $$

$$ = b \left[ (2\sin 2\theta) \cdot \cos 2\theta - (1 - \cos 2\theta) \cdot 2\sin 2\theta \right] $$

$$ = b \left[ 2\sin 2\theta \cos 2\theta - 2\sin 2\theta + 2\sin 2\theta \cos 2\theta \right] $$

$$ = b \left[ 4\sin 2\theta \cos 2\theta - 2\sin 2\theta \right] $$

**Step 3: Calculate $\frac{dy}{dx}$:**

$$ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} $$

Substituting the expressions for $\frac{dy}{d\theta}$ and $\frac{dx}{d\theta}$:

$$ \frac{dy}{dx} = \frac{b \left[ 4\sin 2\theta \cos 2\theta - 2\sin 2\theta \right]}{a \left[ 2\cos 2\theta + 2(\cos^2 2\theta - \sin^2 2\theta) \right]} $$

Simplifying the expression:

$$ = \frac{b \cdot 2\sin 2\theta (2\cos 2\theta - 1)}{a \cdot 2(\cos 2\theta + \cos^2 2\theta - \sin^2 2\theta)} $$

$$ = \frac{b}{a} \cdot \frac{\sin 2\theta (2\cos 2\theta - 1)}{\cos 2\theta + \cos^2 2\theta - \sin^2 2\theta} $$

Recognizing that $\tan \theta = \frac{\sin 2\theta}{\cos 2\theta}$, we see that the expression simplifies to:

$$ \frac{dy}{dx} = \frac{b}{a} \tan \theta $$

Thus, the correct answer is Option 2: $\frac{b}{a} \tan \theta$.

**Explanation of Incorrect Options:**

1. **Option 1: $\frac{b}{a} \cot \theta$** - This option is incorrect because the derivation shows that the expression involves $\tan \theta$, not $\cot \theta$.

3. **Option 3: $\frac{b}{a} \cdot \frac{\sin 2\theta - \cos 2\theta}{\cos \theta}$** - This option is incorrect because the simplification of the derivatives does not lead to this expression.

4. **Option 4: $\frac{a}{b} \sin \theta$** - This option is incorrect because the ratio of derivatives results in a factor of $\frac{b}{a}$, not $\frac{a}{b}$, and involves $\tan \theta$, not $\sin \theta$.

Therefore, the correct answer is indeed Option 2: $\frac{b}{a} \tan \theta$.