Find the eccentricity of the polar equation $$r ( 14 + \sin \theta ) = 28$$
A) 28
B) 14
C) $\frac { 1 } { 28 }$
D) $\frac { 1 } { 14 }$
E) 7

Question

Find the eccentricity of the polar equation $$r ( 14 + \sin \theta ) = 28$$ 
A) 28
B) 14
C) $\frac { 1 } { 28 }$
D) $\frac { 1 } { 14 }$
E) 7

Quizplus · Accepted Answer

Rearrange the equation to get $$r = \frac{28}{14+\sin\theta}.$$ Then, from the definition of eccentricity, $e = \frac{c}{a}$, where $c$ is the distance from the origin to the focus and $a$ is the distance from the origin to the directrix. Since the polar equation is of the form $r = \frac{L}{1-e\cos\theta}$, it follows that $c = eL$ and $a = \frac{L}{1-e^2}$. To find $e$, we need to determine $L$, which is the distance from the origin to the directrix.

One way to find $L$ is to note that the directrix is perpendicular to the line passing through the focus and the pole that makes an angle of $\frac{\pi}{2}$ with the polar axis. This line is given by $y = -x\tan\left(\frac{\pi}{2}-\frac{\theta}{2}\right) = x\cot\left(\frac{\theta}{2}\right)$. Since the directrix is the line that is equidistant from the pole and the focus, we have $$r = r_0 \pm \frac{L}{2},$$ where $r_0$ is the distance from the pole to the directrix, $\frac{L}{2}$ is the distance from the focus to the directrix, and the sign depends on whether the directrix is above or below the pole. Substituting $r = \frac{28}{14+\sin\theta}$ and solving for $\sin\theta$, we get $$\sin\theta = \frac{28r-392}{r+196}.$$ Substituting this in the equation for the line, we get $$r = \frac{x}{\cos\theta} = \frac{y}{\sin\theta} = \frac{x}{\cos\theta}\cdot\frac{r+196}{28r-392},$$ which simplifies to $$x = \frac{196}{\cos\theta-2}.$$ Therefore, $L = \frac{196}{2} = 98$, so $$e = \frac{c}{a} = \frac{L}{2r_0} = \frac{98}{2\cdot 14} = \boxed{\frac{1}{14}}.$$

Find the Eccentricity of the Polar Equation r(14+sin⁡θ)=28r ( 14 + \sin \theta ) = 28r(14+sinθ)=28

Find the Eccentricity of the Polar Equation $r ( 14 + \sin \theta ) = 28$