# (a) Given the conic section displaystyle{r}=frac{5}{{{7}+{3} cos{{left(thetaright)}}}}, find the x and y intercept(s) and the focus(foci). (b) Given the conic section displaystyle{r}=frac{5}{{{2}+{5} sin{{left(thetaright)}}}}, find the x and y intercept(s) and the focus(foci).

Carol Gates 2020-12-16 Answered
(a) Given the conic section $r=\frac{5}{7+3\mathrm{cos}\left(\theta \right)}$, find the x and y intercept(s) and the focus(foci).
(b) Given the conic section $r=\frac{5}{2+5\mathrm{sin}\left(\theta \right)}$, find the x and y intercept(s) and the focus(foci).
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Nola Robson
(a)
Given $r=\frac{5}{7+3\mathrm{cos}\theta }$
It can be written as $r=\frac{5\text{/}7}{1+3\text{/}7\mathrm{cos}\theta }$
Comparing it with general conic section $r=\frac{ed}{1+e\mathrm{cos}\theta }$, we get
$e=\frac{3}{7}$, since e i.e. eccentricity lies between 0 & 1. :. given conic section is ellipse.
for x-intercept $\theta =0$,
so x-intercept in this case is (1/2, 0)
when $\theta =\pi .r=\frac{5}{4}$, so x-intercept in this case is $\left(-\frac{5}{4},0\right)$
for y-intercept $\theta =\frac{\pi }{2},\frac{3\pi }{2}$
when $\theta =\frac{\pi }{2},r=\frac{5}{7}$, so y-intercept in this case is $\left(0,\frac{5}{7}\right)$
when $\theta =\frac{3\pi }{2},r=\frac{5}{7}$, so y-intercept in this case is $\left(0,-\frac{5}{7}\right)$
One foci is (0, 0) of this form. Now, to find other foci we will find out center of the ellipse & it will be the mid point of both the x-intercepts
i.e. center $=\left(\frac{\frac{1}{2}-\frac{5}{4}}{2},0\right)$
$=\left(-\frac{3}{8},0\right)$
Now we will distance of center $\left(-\frac{3}{8},0\right)$ and 1st foci $\left(0,0\right),c=3/8$
since the distance of both the foci from the center is same, so coordinates of 2nd foci is $\left(-\frac{3}{8}-\frac{3}{8},0\right)$
i.e. $\left(-\frac{6}{8},0\right)$
so, x intercepts are $\left(\frac{1}{2},0\right),\left(-\frac{5}{4},0\right)$
y intercepts are $\left(0,\frac{5}{7}\right),\left(0,-\frac{5}{7}\right)$
foci are $\left(0,0\right),\left(-\frac{6}{8},0\right)$
(b)
Given $r=\frac{5}{2+5\mathrm{sin}\theta }$
It can be written as $r=\frac{5\text{/}2}{1+5\text{/}2\mathrm{sin}\theta }$
Comparing it with general conic section $r=\frac{ed}{1+e\mathrm{sin}\theta }$, we get
$e=\frac{5}{2}$, since e i.e. eccentricity is greater than $1.\therefore$ given conic section is hyperbola.
for x-intercept $\theta =0,\pi$
when $\theta =0,r=\frac{5}{2}$, so x -intercept in this case is $\left(\frac{5}{2},0\right)$
when $\theta =\pi .r=\frac{5}{2}$,so x -intercept in this case is $\left(-\frac{5}{2},0\right)$
for y-intercept $\theta =\frac{\pi }{2},\frac{3\pi }{2}$
when $\theta =\frac{\pi }{2},r=\frac{5}{7}$, so y-intercept in this case is $\left(0,\frac{5}{7}\right)$
when