Angle of intersection of two curves
Problem (a): Find the angle of intersection of the curves \( y = 6 - x^2 \) and \( x^2 = 2y \).
1. Intersection points
Substitute \(y = 6 - x^2\) into \(x^2 = 2y\): \[ x^2 = 2(6 - x^2) \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2. \] So the curves meet at \((2,2)\) and \((-2,2)\).
Substitute \(y = 6 - x^2\) into \(x^2 = 2y\): \[ x^2 = 2(6 - x^2) \Rightarrow 3x^2 = 12 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2. \] So the curves meet at \((2,2)\) and \((-2,2)\).
2. Slopes (dy/dx)
For \(y = 6 - x^2\): \(\dfrac{dy}{dx} = -2x\).
For \(x^2 = 2y\): differentiate \(2x = 2\dfrac{dy}{dx}\) so \(\dfrac{dy}{dx} = x\).
At \(x=2\): slopes are \(m_1=-4\) and \(m_2=2\). At \(x=-2\): slopes are \(m_1=4\) and \(m_2=-2\).
For \(y = 6 - x^2\): \(\dfrac{dy}{dx} = -2x\).
For \(x^2 = 2y\): differentiate \(2x = 2\dfrac{dy}{dx}\) so \(\dfrac{dy}{dx} = x\).
At \(x=2\): slopes are \(m_1=-4\) and \(m_2=2\). At \(x=-2\): slopes are \(m_1=4\) and \(m_2=-2\).
3. Angle between tangents
Use \[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|. \] At \((2,2)\): \[ \tan\theta = \left|\frac{-4 - 2}{1 + (-4)(2)}\right| = \left|\frac{-6}{1 - 8}\right| = \frac{6}{7}. \] At \((-2,2)\) the same value appears (symmetry).
Use \[ \tan\theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|. \] At \((2,2)\): \[ \tan\theta = \left|\frac{-4 - 2}{1 + (-4)(2)}\right| = \left|\frac{-6}{1 - 8}\right| = \frac{6}{7}. \] At \((-2,2)\) the same value appears (symmetry).
Exact angle:
\[
\theta = \tan^{-1}\!\left(\frac{6}{7}\right).
\]
Numeric (degrees):
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You may copy the exact result \(\tan^{-1}\!\frac{6}{7}\) into any CAS or calculator. The numeric value in degrees is shown above.