Find `dy/dx`, if x and y are connected parametrically by the equations, given below without eliminating the parameter:
`x=atan^(2)theta,y=bsec^(2)theta`

Answer

To find \(\frac{dy}{dx}\) given the parametric equations \(x = a \tan^2 \theta\) and \(y = b \sec^2 \theta\), we will differentiate both equations with respect to \(\theta\) and then use the chain rule.

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

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

Using the chain rule:
\[ \frac{dx}{d\theta} = a \cdot 2 \tan \theta \cdot \frac{d}{d\theta}(\tan \theta) = a \cdot 2 \tan \theta \cdot \sec^2 \theta \]
So,
\[ \frac{dx}{d\theta} = 2a \tan \theta \sec^2 \theta \]

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

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

Using the chain rule:
\[ \frac{dy}{d\theta} = b \cdot 2 \sec \theta \cdot \frac{d}{d\theta}(\sec \theta) = b \cdot 2 \sec \theta \cdot \sec \theta \tan \theta \]
So,
\[ \frac{dy}{d\theta} = 2b \sec^2 \theta \tan \theta \]

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

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

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{2b \sec^2 \theta \tan \theta}{2a \tan \theta \sec^2 \theta} \]

### Step 4: Simplify the expression

Notice that \(2\), \(\sec^2 \theta\), and \(\tan \theta\) in the numerator and denominator can be simplified:
\[ \frac{dy}{dx} = \frac{2b \sec^2 \theta \tan \theta}{2a \tan \theta \sec^2 \theta} = \frac{b}{a} \]

Thus, the final result is:
\[ \frac{dy}{dx} = \frac{b}{a} \]