Question

The angle of intersection of the curve $y = x^2$ and $x = y^2$ in the first quadrant is $\tan^{-1}(\frac{p}{q})$ where $p, q$ are co-prime natural numbers then $p + q$ is _____.

Answer 1

7

Answer 2

To find the angle of intersection of two curves, we first need to find their points of intersection. Then, at each point of intersection, we calculate the slopes of the tangents to both curves. Finally, we use the formula for the angle between two lines.

1. Find the points of intersection: The given curves are: Curve 1: $y = x^2$ Curve 2: $x = y^2$

Substitute the expression for $y$ from Curve 1 into Curve 2: $x = (x^2)^2$ $x = x^4$ $x^4 - x = 0$ $x(x^3 - 1) = 0$

This equation gives two possible values for $x$:

• $x = 0$
• $x^3 - 1 = 0 \Rightarrow x^3 = 1 \Rightarrow x = 1$ (since we are looking for real intersections)

Now find the corresponding $y$ values using $y = x^2$:

• If $x = 0$, then $y = 0^2 = 0$. So, $(0,0)$ is an intersection point.
• If $x = 1$, then $y = 1^2 = 1$. So, $(1,1)$ is an intersection point.

Both $(0,0)$ and $(1,1)$ are in the first quadrant (or on its boundary).

2. Calculate the slopes of the tangents: We need to find $\frac{dy}{dx}$ for each curve.

For Curve 1: $y = x^2$ $\frac{dy}{dx} = 2x$

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

3. Calculate the angle of intersection at each point:

At the point $(0,0)$:

• Slope of tangent to Curve 1 ($m_1$): $m_1 = 2(0) = 0$. (The tangent is the x-axis)
• Slope of tangent to Curve 2 ($m_2$): $m_2 = \frac{1}{2(0)}$, which is undefined. (The tangent is the y-axis) When one slope is $0$ and the other is undefined, the angle between the tangents is $90^\circ$. $\tan 90^\circ$ is undefined. The form $\tan^{-1}(\frac{p}{q})$ with $p, q$ being natural numbers implies a finite value, so this point is likely not what the question is looking for.

At the point $(1,1)$:

• Slope of tangent to Curve 1 ($m_1$): $m_1 = 2(1) = 2$.
• Slope of tangent to Curve 2 ($m_2$): $m_2 = \frac{1}{2(1)} = \frac{1}{2}$.

The angle $\theta$ between two curves with slopes $m_1$ and $m_2$ is given by: $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ $\tan \theta = \left| \frac{2 - \frac{1}{2}}{1 + (2)\left(\frac{1}{2}\right)} \right|$ $\tan \theta = \left| \frac{\frac{4-1}{2}}{1 + 1} \right|$ $\tan \theta = \left| \frac{\frac{3}{2}}{2} \right|$ $\tan \theta = \left| \frac{3}{4} \right|$ $\tan \theta = \frac{3}{4}$

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

4. Determine $p+q$: The angle is given in the form $\tan^{-1}(\frac{p}{q})$, where $p, q$ are co-prime natural numbers. Comparing $\tan^{-1}\left(\frac{3}{4}\right)$ with $\tan^{-1}\left(\frac{p}{q}\right)$, we get $p=3$ and $q=4$. $p=3$ and $q=4$ are natural numbers. They are co-prime because their greatest common divisor is 1. Finally, we need to find $p+q$: $p+q = 3+4 = 7$.