Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=4+2 \cos \theta, \quad y=-1+\sin \theta $$

Short answer

The points of vertical tangency to the curve are (6, -1) and (2, -1). The points of horizontal tangency to the curve are (4, 0) and (4, -2).

Step-by-step solution

Define the derivatives

Begin by defining the derivatives dx/dθ and dy/dθ from the given parametric equations, so\(dx/dθ = -2 \sin \theta\) and \(dy/dθ = \cos \theta\)

Find the points of vertical tangency

Vertical tangency occurs when the derivative in x is zero or undefined. Set dx/dθ to 0 and solve for θ:\(-2 \sin \theta = 0\)Therefore \(\theta = 0, \pi\)Substitute these θ values in the original equations for x and y to get the points of vertical tangency: For \(\theta = 0\), \(x = 4 + 2(1)\), \(y = -1 + 0\) which gives (6, -1)For \(\theta = \pi\), \(x = 4 + 2(-1)\), \(y = -1 + 0\) which gives (2, -1)

Find the points of horizontal tangency

Horizontal tangency occurs when the derivative in y is zero or undefined. Set dy/dθ to 0 and solve for θ:\(\cos \theta = 0\)Therefore \(\theta = \pi/2, (3\pi)/2\)Substitute these θ values in the original equations for x and y to get the points of horizontal tangency: For \(\theta = \pi/2\), \(x = 4 + 2(0)\), \(y = -1 + 1\) which gives (4,0)For \(\theta = (3\pi)/2\), \(x = 4 - 2(0)\), \(y = -1 - 1\) which gives (4, -2)

Confirm your results

The final step is to confirm these results using a graphing utility. Plot the curve and see if it crosses the y-axis at the points found for vertical tangency and likewise for the x-axis and horizontal tangency. The curve should be tangent to these points.