If `x=asec^(2)theta,y=btan^(2)theta," then "(dy)/(dx)=`

Options

A. `((a)/(b))csctheta`
B. `(-(a)/(b))cottheta`
C. 1
D. `((b)/(a))`

Answer

To find \(\frac{dy}{dx}\) given \(x = a \sec^2 \theta\) and \(y = b \tan^2 \theta\), we will use the chain rule of differentiation. Here’s the step-by-step solution:

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

Given:
\[ x = a \sec^2 \theta \]

Differentiating both sides with respect to \(\theta\):
\[ \frac{dx}{d\theta} = a \cdot \frac{d}{d\theta}(\sec^2 \theta) \]

Using the derivative of \(\sec^2 \theta\):
\[ \frac{d}{d\theta}(\sec^2 \theta) = 2 \sec^2 \theta \tan \theta \]

Thus,
\[ \frac{dx}{d\theta} = a \cdot 2 \sec^2 \theta \tan \theta = 2a \sec^2 \theta \tan \theta \]

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

Given:
\[ y = b \tan^2 \theta \]

Differentiating both sides with respect to \(\theta\):
\[ \frac{dy}{d\theta} = b \cdot \frac{d}{d\theta}(\tan^2 \theta) \]

Using the derivative of \(\tan^2 \theta\):
\[ \frac{d}{d\theta}(\tan^2 \theta) = 2 \tan \theta \sec^2 \theta \]

Thus,
\[ \frac{dy}{d\theta} = b \cdot 2 \tan \theta \sec^2 \theta = 2b \tan \theta \sec^2 \theta \]

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

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]

Substituting the results from Steps 1 and 2:
\[ \frac{dy}{dx} = \frac{2b \tan \theta \sec^2 \theta}{2a \sec^2 \theta \tan \theta} \]

### Step 4: Simplify the expression

The \(\tan \theta\) and \(\sec^2 \theta\) terms cancel out:
\[ \frac{dy}{dx} = \frac{b}{a} \]

### Final Answer
\[ \frac{dy}{dx} = \frac{b}{a} \]

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