# Please help How do you graph the polar equation r=8-sectheta ?

Aug 10, 2018

See the explanation and graphs.

#### Explanation:

As sec values $\notin \left(- 1 , 1\right)$,

$r = 8 - \sec \theta \notin \left(- 1 + 8 , 1 + 8\right) = \left(7 , 9\right)$

r is periodic,with period $2 \pi$.

Short Table, for $theta in $\theta \in \left[0 , \pi\right]$, sans asymptotic $\frac{\pi}{2} \mathmr{and} \frac{3}{2} \pi$: $\left(r , \theta\right)$: ( 7, 0 ) ( 6.845, pi/6 ) ( 6, pi/3 ) ( oo, pi/2 ) $\left(9.155 , \frac{5}{6} \pi\right) \left(10 , 2 \frac{\pi}{3}\right) \left(9 , \pi\right)$. $r = 0 , a t \theta = 1.4455$rad. The graph is symmetrical about $\theta = 0$. graph is outside the annular ( circular ) region  7 < r < 9 ). Converting to the Cartesian form, using $\left(x , y\right) = r \left(\cos \theta , \sin \theta\right) \mathmr{and} 0 \le \sqrt{{x}^{2} + {y}^{2}} = r$, ${\left({x}^{2} + {y}^{2}\right)}^{0.5} \left(1 + \frac{1}{x}\right) = 8$, The Socratic graph is immediate, with asymptote x = 0. graph{((x^2 + y^2)^0.5(1+1/x) - 8)(x+0.01y)=0[-40 40 -20 20]} See the bounding circles $r = 7 \mathmr{and} r = 9\$.
graph{((x^2 + y^2)^0.5(1+1/x) - 8)(x^2+y^2-49)(x^2+y^2-81)=0[-20 20 -10 10]}