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: \( \boxed{\tan^{-1}(3)} \).

Answer 2

Step 1: Find the point of intersection

From the first curve: \[ y^2 = 2x \Rightarrow x = \frac{y^2}{2} \] Substitute into the second curve: \[ \left(\frac{y^2}{2}\right)^2 + y^2 = 8 \Rightarrow \frac{y^4}{4} + y^2 = 8 \Rightarrow y^4 + 4y^2 = 32 \Rightarrow y^4 + 4y^2 - 32 = 0 \] Let \( z = y^2 \), then: \[ z^2 + 4z - 32 = 0 \Rightarrow z = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm \sqrt{144}}{2} = \frac{-4 \pm 12}{2} \Rightarrow z = 4 \text{ or } -8 \] Discard \( z = -8 \) since \( y^2 \ge 0 \). So \( y^2 = 4 \Rightarrow y = \pm 2 \) Now: \[ x = \frac{y^2}{2} = \frac{4}{2} = 2 \] So points of intersection: \( (2, 2) \) and \( (2, -2) \)

Step 2: Find the slopes at point of intersection

For \( y^2 = 2x \), differentiate: \[ 2y \frac{dy}{dx} = 2 \Rightarrow \frac{dy}{dx} = \frac{1}{y} \] At \( (2, 2) \): slope = \( \frac{1}{2} \) For \( x^2 + y^2 = 8 \), differentiate: \[ 2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y} \] At \( (2, 2) \): slope = \( -\frac{2}{2} = -1 \)

Step 3: Use the formula for angle between curves

Let \( m_1 = \frac{1}{2} \), \( m_2 = -1 \) The angle \( \theta \) between curves is given by: \[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{\frac{1}{2} - (-1)}{1 + \frac{1}{2} \cdot (-1)} \right| = \left| \frac{\frac{3}{2}}{1 - \frac{1}{2}} \right| = \left| \frac{\frac{3}{2}}{\frac{1}{2}} \right| = 3 \Rightarrow \theta = \tan^{-1}(3) \]

Answer:

\( \boxed{\tan^{-1}(3)} \)

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