Tangents with Parametric Equations — Question 7

Question 7

Problem

Find the angle at which the parametric curve $x=2\cos t$, $y=\sin t$ crosses the line $y=x/2$ in the first quadrant.

Question 7 – Solution

Solution

1. At an intersection with $y=x/2$, substitute the parametric equations: $$\sin t=\frac{2\cos t}{2}=\cos t.$$ In the first quadrant, this gives $t=\pi/4$.

2. The intersection point is $$\left(2\cos\frac\pi4,\sin\frac\pi4\right) =\left(\sqrt2,\frac{\sqrt2}{2}\right).$$

3. Differentiate the curve: $$\frac{dx}{dt}=-2\sin t, \qquad \frac{dy}{dt}=\cos t.$$ Thus its tangent slope at $t=\pi/4$ is $$m_c=\frac{\cos t}{-2\sin t}=-\frac12.$$ The line $y=x/2$ has slope $m_\ell=1/2$.

4. If $\phi$ is the acute angle between two lines, then $$\tan\phi=\left|\frac{m_\ell-m_c}{1+m_\ell m_c}\right|.$$ Therefore, $$\tan\phi =\left|\frac{\frac12-(-\frac12)}{1+(\frac12)(-\frac12)}\right| =\frac{1}{3/4} =\frac43.$$

5. Hence the crossing angle is $$\boxed{\phi=\arctan\left(\frac43\right)\approx53.13^\circ}.$$