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 between the curves \(y = x^2\) and \(x = y^2\) at the point (1,1), we can use the formula for the angle between two curves:

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

\[\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|\]

First, we find the derivatives of the curves to get the slopes:

1. Differentiating \(y = x^2\) with respect to \(x\), we get: 

\[\frac{dy}{dx} = 2x\]

1. . At the point (1,1), the slope \(m_1 = 2(1) = 2\).
2. Differentiating \(x = y^2\) with respect to \(x\), we implicitly differentiate to get: 

\[\frac{d}{dx}(x = y^2) \Rightarrow 1 = 2y \frac{dy}{dx}\]

1. 
   Solving for \(\frac{dy}{dx}\), we have: 

\[\frac{dy}{dx} = \frac{1}{2y}\]

1. . At the point (1,1), the slope \(m_2 = \frac{1}{2(1)} = \frac{1}{2}\).

Substituting the values of \(m_1\) and \(m_2\) into the formula for \(\tan \theta\), we get:

\[\tan \theta = \left|\frac{2 - \frac{1}{2}}{1 + 2 \times \frac{1}{2}}\right|\]

Calculate the expression:

\[\tan \theta = \left|\frac{\frac{4}{2} - \frac{1}{2}}{1 + 1}\right| = \left|\frac{\frac{3}{2}}{2}\right| = \left|\frac{3}{4}\right|\]

Thus, the angle of intersection \(\theta\) is \(\tan^{-1}\left(\frac{3}{4}\right)\).

The correct answer is:

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