Question

The angle between the curve $y^2=x$ and $x^2=y$ at (1, 1) is

• A. 60°
• B. tan-143
• C. cot-143
• D. 90°

Answer 1

Answer: (C)

cot-143

Answer 2

To find the angle between two curves at their intersection point, we first find the slopes of their tangents at that point and then use the formula for the angle between two lines.

1. Identify the curves and the intersection point:

   Curve 1: $y^2 = x$ Curve 2: $x^2 = y$ Intersection point: (1, 1) (Verify by substituting: $1^2=1$ for both equations)

2. Find the slope of the tangent to Curve 1 at (1, 1):

   Differentiate $y^2 = x$ with respect to $x$: $2y \frac{dy}{dx} = 1$ $\frac{dy}{dx} = \frac{1}{2y}$

   Let $m_1$ be the slope of the tangent to Curve 1 at (1, 1): $m_1 = \left(\frac{dy}{dx}\right)_{(1,1)} = \frac{1}{2(1)} = \frac{1}{2}$

3. Find the slope of the tangent to Curve 2 at (1, 1):

   Differentiate $x^2 = y$ with respect to $x$: $2x = \frac{dy}{dx}$

   Let $m_2$ be the slope of the tangent to Curve 2 at (1, 1): $m_2 = \left(\frac{dy}{dx}\right)_{(1,1)} = 2(1) = 2$

4. Calculate the angle between the tangents:

   The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:

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

   Substitute $m_1 = \frac{1}{2}$ and $m_2 = 2$:

   $$\tan \theta = \left|\frac{2 - \frac{1}{2}}{1 + \left(\frac{1}{2}\right)(2)}\right| = \left|\frac{\frac{3}{2}}{2}\right| = \left|\frac{3}{4}\right| = \frac{3}{4}$$

   From this, we get $\theta = \tan^{-1}\left(\frac{3}{4}\right)$.

5. Match with the given options:

   We found $\tan \theta = \frac{3}{4}$. This means $\cot \theta = \frac{1}{\tan \theta} = \frac{1}{3/4} = \frac{4}{3}$. Therefore, $\theta = \cot^{-1}\left(\frac{4}{3}\right)$.

   The answer is $\cot^{-1}\left(\frac{4}{3}\right)$, which corresponds to option C, assuming "cot-143" is a typo and should be interpreted as $\cot^{-1}\left(\frac{4}{3}\right)$.