Question

The angle between the curves \( y^2 = 2x \) and \( x^2 + y^2 = 8 \) is 

• A. \( \tan^{-1}(1) \)
• B. \( \tan^{-1}(2) \)
• C. \( \tan^{-1}(3) \)
• D. \( \tan^{-1}\left(-\frac{1}{2}\right) \)

Hint:

To find the angle between two curves, differentiate both equations to find their slopes, then apply the angle formula for intersecting curves.

Answer 1

The Correct Option is C

Step 1: Find the slopes of the given curves 
The given equations of the curves are: 1. \( y^2 = 2x \) 2. \( x^2 + y^2 = 8 \) 
Step 2: Find the derivative for the first curve 
Differentiating \( y^2 = 2x \) with respect to \( x \), using implicit differentiation: \[ 2y \frac{dy}{dx} = 2 \] \[ \frac{dy}{dx} = \frac{1}{y}. \] 
Step 3: Find the derivative for the second curve 
Differentiating \( x^2 + y^2 = 8 \) with respect to \( x \): \[ 2x + 2y \frac{dy}{dx} = 0. \] \[ \frac{dy}{dx} = -\frac{x}{y}. \] 
Step 4: Compute the angle between the two curves 
The angle \( \theta \) between the two curves is given by the formula: \[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|, \] where \( m_1 = \frac{1}{y} \) and \( m_2 = -\frac{x}{y} \). Substituting these values: \[ \tan \theta = \left| \frac{\frac{1}{y} + \frac{x}{y}}{1 - \frac{x}{y} \cdot \frac{1}{y}} \right|. \] Simplifying, we get: \[ \tan \theta = 3. \] 
Final Answer: \( \boxed{\tan^{-1}(3)} \).