Find all points (if any) of horizontal and vertical tangency to the curve $x = 7 + 3 \cos \theta , y = - 3 + \sin \theta$
A) horizontal tangents: $( 7 , - 2 ) , ( 7 , - 4 )$, vertical tangents: $( 10 , - 3 ) , ( 4 , - 3 )$
B) horizontal tangents: $( 10 , - 3 ) , ( 4 , - 3 )$, vertical tangents: $( 7 , - 2 ) , ( 7 , - 4 )$
C) horizontal tangent: $( 7 , - 2 )$, vertical tangent: $( 10 , - 3 )$
D) horizontal tangents: $( 7 , - 4 ) , ( 7 , - 4 )$, vertical tangents: $( 10 , - 3 ) , ( 4 , - 3 )$
E) horizontal tangent: $( 10 , - 3 )$, vertical tangent: $( 7 , - 2 )$

Question

Quizplus · Accepted Answer

Horizontal tangency occurs when the derivative of $y$ with respect to $x$ ($dy/dx$) is zero, and vertical tangency occurs when $dy/dx$ is undefined or infinite. To find $dy/dx$, we first find $dx/d\theta = -3\sin\theta$ and $dy/d\theta = \cos\theta$. Thus, $dy/dx = (dy/d\theta)/(dx/d\theta) = (\cos\theta)/(-3\sin\theta) = -\cot\theta/3$. For horizontal tangents, we set $dy/dx = 0$, which occurs when $\sin\theta = 0$, leading to $\theta = 0, \pi$. Substituting these values into the original equations for $x$ and $y$, we get points $(10, -3)$ and $(4, -3)$ for vertical tangents. For vertical tangents, $dy/dx$ is undefined when $\cos\theta = 0$, leading to $\theta = \pi/2, 3\pi/2$. Substituting these into the original equations gives $(7, -2)$ and $(7, -4)$ for horizontal tangents.

Find All Points (If Any) of Horizontal and Vertical Tangency x=7+3cos⁡θ,y=−3+sin⁡θx = 7 + 3 \cos \theta , y = - 3 + \sin \thetax=7+3cosθ,y=−3+sinθ

Find All Points (If Any) of Horizontal and Vertical Tangency $x = 7 + 3 \cos \theta , y = - 3 + \sin \theta$