Question

Find all points (if any) of horizontal and vertical tangency to the curve. Use a graphing utility to confirm your results. $$ x=\sec \theta, \quad y=\tan \theta $$

Short answer

The point of horizontal tangency to the curve is at (1,0). There are no points of vertical tangency to the curve.

Step-by-step solution

Convert parametric to rectangular

Write the expression for \( x \) in terms of \( y \). Using the trigonometric identity \( \sec^{2}(\theta)=1+\tan^{2}(\theta) \), replace \( \sec \theta \) in the equation \( x=\sec \theta \) by \( x=\sqrt{1+y^{2}} \). Thus, the curve is described by the equation \( x=\sqrt{1+y^{2}} \).

Find the derivative

Next, calculate the derivative of \( x \) in terms of \( y \). The derivative of \( x \) with respect to \( y \) is \( dx/dy=y/(y^{2}+1) \). Keep this expression for the subsequent steps.

Find points of horizontal tangency

Horizontal tangency points occur where \( dx/dy=0 \). Hence, solve the equation \( y/(y^{2}+1)=0 \). The solution for \( y \) is \( y=0 \). After substituting \( y=0 \) in the rectangular equation \( x=\sqrt{1+y^{2}} \), the corresponding \( x \) value is \( x=1 \). Hence, the point of horizontal tangency is (1,0).

Find points of vertical tangency

Vertical tangency occurs where the derivative is undefined. Hence, solve the equation \( y^{2}+1=0 \). However, there is no solution to this equation in the real number system. Hence, there are no points of vertical tangency to the curve.

Validate with graphing utility

After plotting the curve \( x=\sqrt{1+y^{2}} \) using a graphing utility, it can be seen that the curve indeed has a flat point at (1,0), confirming our above analysis.