Question

the angle of intersection of the curves y=x² and x=y² at (1,1) is

• A. tan^-1\(\frac{4}{3}\)
• B. tan^-1(1)
• C. 90\(^{\circ}\)
• D. tan^-1\(\frac{3}{4}\)

Answer 1

The Correct Option is D

To find the angle of intersection of the curves \( y = x^2 \) and \( x = y^2 \) at the point \( (1, 1) \), we need to follow these steps:

1. Find the derivatives:
   • For the curve \( y = x^2 \), the derivative is \(\frac{dy}{dx} = 2x\).
   • For the curve \( x = y^2 \), we can differentiate implicitly to find \(\frac{dy}{dx}\). Differentiating both sides with respect to \(x\) gives:
     \[ 1 = 2y \cdot \frac{dy}{dx} \] Therefore, \[ \frac{dy}{dx} = \frac{1}{2y} \]
2. Calculate slopes at the point (1,1):
   • For \( y = x^2 \), substitute \( x = 1 \): \[ \frac{dy}{dx} = 2 \cdot 1 = 2 \]
   • For \( x = y^2 \), substitute \( y = 1 \): \[ \frac{dy}{dx} = \frac{1}{2 \cdot 1} = \frac{1}{2} \]
3. Find the angle of intersection:

   The angle \(\theta\) between two curves with slopes \(m_1\) and \(m_2\) is given by the formula:

   \[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1m_2}\right| \]
   • Substitute \( m_1 = 2 \) and \( m_2 = \frac{1}{2} \): \[ \tan\theta = \left|\frac{2 - \frac{1}{2}}{1 + 2 \cdot \frac{1}{2}}\right| = \left|\frac{\frac{4}{2} - \frac{1}{2}}{1 + 1}\right| = \left|\frac{\frac{3}{2}}{2}\right| = \left|\frac{3}{4}\right| \]

Therefore, the angle \(\theta\) can be expressed as \(\tan^{-1}\left(\frac{3}{4}\right)\).

Conclusion: The angle of intersection of the curves at the point (1,1) is \(\tan^{-1}\left(\frac{3}{4}\right)\). Thus, the correct answer is:

tan^-1\(\frac{3}{4}\)