If ` x =a sqrt( sectheta -tan theta) ,y=a sqrt(sectheta +tan theta ), then (dy)/(dx) =`

Options

A. ` (y)/(x)`
B. ` (-y)/(x)`
C. ` (x)/(y)`
D. ` (-x)/(y)`

Answer

To solve the problem, we need to find the derivative \( \frac{dy}{dx} \) given the equations:

\[ x = a \sqrt{\sec \theta - \tan \theta} \]
\[ y = a \sqrt{\sec \theta + \tan \theta} \]

### Step 1: Square both equations

First, we square both \( x \) and \( y \):

\[ x^2 = a^2 (\sec \theta - \tan \theta) \]
\[ y^2 = a^2 (\sec \theta + \tan \theta) \]

### Step 2: Express \( \sec \theta \) and \( \tan \theta \)

From the above equations, we can express \( \sec \theta \) and \( \tan \theta \):

\[ \sec \theta = \frac{x^2}{a^2} + \tan \theta \]
\[ \sec \theta = \frac{y^2}{a^2} - \tan \theta \]

### Step 3: Add and subtract the equations

Now, we can add and subtract these two equations to eliminate \( \tan \theta \):

Adding:
\[ \sec \theta + \sec \theta = \frac{x^2}{a^2} + \frac{y^2}{a^2} \]
\[ 2\sec \theta = \frac{x^2 + y^2}{a^2} \]
\[ \sec \theta = \frac{x^2 + y^2}{2a^2} \]

Subtracting:
\[ \sec \theta - \sec \theta = \frac{x^2}{a^2} - \frac{y^2}{a^2} + 2\tan \theta \]
\[ 0 = \frac{x^2 - y^2}{a^2} + 2\tan \theta \]
\[ \tan \theta = -\frac{x^2 - y^2}{2a^2} \]

### Step 4: Use the identity for \( \sec \theta \) and \( \tan \theta \)

Using the identity \( \sec^2 \theta = 1 + \tan^2 \theta \):

\[ \sec^2 \theta = \left(\frac{x^2 + y^2}{2a^2}\right)^2 + \left(-\frac{x^2 - y^2}{2a^2}\right)^2 \]

### Step 5: Differentiate implicitly

Now we differentiate the equation \( x^2y^2 = a^4 \) implicitly with respect to \( x \):

Using the product rule:
\[ \frac{d}{dx}(x^2y^2) = \frac{d}{dx}(a^4) \]
\[ 2xy^2 + 2x^2\frac{dy}{dx} = 0 \]

### Step 6: Solve for \( \frac{dy}{dx} \)

Rearranging the equation:
\[ 2x^2\frac{dy}{dx} = -2xy^2 \]
\[ \frac{dy}{dx} = -\frac{xy^2}{x^2} \]

### Final Result

Thus, we have:

\[ \frac{dy}{dx} = -\frac{y}{x} \]