`x=2 cos^2 t, y= 6 sin ^2 t `

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = 2 \cos^2 t\) and \(y = 6 \sin^2 t\), we will follow these steps:

### Step 1: Differentiate \(x\) with respect to \(t\)

Given:
\[ x = 2 \cos^2 t \]

Using the chain rule, we differentiate \(x\) with respect to \(t\):
\[ \frac{dx}{dt} = 2 \cdot 2 \cos t \cdot (-\sin t) = -4 \cos t \sin t \]

### Step 2: Differentiate \(y\) with respect to \(t\)

Given:
\[ y = 6 \sin^2 t \]

Using the chain rule, we differentiate \(y\) with respect to \(t\):
\[ \frac{dy}{dt} = 6 \cdot 2 \sin t \cdot \cos t = 12 \sin t \cos t \]

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

Using the formula for derivatives of parametric equations:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{12 \sin t \cos t}{-4 \cos t \sin t} \]

### Step 4: Simplify the expression

Notice that \(\sin t\) and \(\cos t\) can be canceled out (assuming \(\sin t \neq 0\) and \(\cos t \neq 0\)):
\[ \frac{dy}{dx} = \frac{12}{-4} = -3 \]

### Final Answer

Thus, the value of \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = -3 \]

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