(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).

(a)
Given ${\displaystyle r=\frac{5}{7+3\cos \theta }}$
It can be written as ${\displaystyle r=\frac{5\text{/}7}{1+3\text{/}7\cos \theta }}$
Comparing it with general conic section ${\displaystyle r=\frac{ed}{1+e\cos \theta }}$, we get
${\displaystyle e=\frac{3}{7}}$, since e i.e. eccentricity lies between 0 & 1. :. given conic section is ellipse.
for x-intercept ${\displaystyle \theta =0}$,
${\displaystyle \pi when\theta =0,r=\frac{5}{10}=\frac{1}{2},}$ so x-intercept in this case is (1/2, 0)
when ${\displaystyle \theta =\pi .r=\frac{5}{4}}$, so x-intercept in this case is ${\displaystyle (-\frac{5}{4},0)}$
for y-intercept ${\displaystyle \theta =\frac{\pi }{2},\frac{3\pi }{2}}$
when ${\displaystyle \theta =\frac{\pi }{2},r=\frac{5}{7}}$, so y-intercept in this case is ${\displaystyle (0,\frac{5}{7})}$
when ${\displaystyle \theta =\frac{3\pi }{2},r=\frac{5}{7}}$, so y-intercept in this case is ${\displaystyle (0,-\frac{5}{7})}$
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 ${\displaystyle =(\frac{\frac{1}{2}-\frac{5}{4}}{2},0)}$
${\displaystyle =(-\frac{3}{8},0)}$
Now we will distance of center ${\displaystyle (-\frac{3}{8},0)}$ and 1st foci $(0,0),c=3/8$
since the distance of both the foci from the center is same, so coordinates of 2nd foci is ${\displaystyle (-\frac{3}{8}-\frac{3}{8},0)}$
i.e. ${\displaystyle (-\frac{6}{8},0)}$
so, x intercepts are ${\displaystyle (\frac{1}{2},0),(-\frac{5}{4},0)}$
y intercepts are ${\displaystyle (0,\frac{5}{7}),(0,-\frac{5}{7})}$
foci are ${\displaystyle (0,0),(-\frac{6}{8},0)}$
(b)
Given ${\displaystyle r=\frac{5}{2+5\sin \theta }}$
It can be written as ${\displaystyle r=\frac{5\text{/}2}{1+5\text{/}2\sin \theta }}$
Comparing it with general conic section ${\displaystyle r=\frac{ed}{1+e\sin \theta }}$, we get
${\displaystyle e=\frac{5}{2}}$, since e i.e. eccentricity is greater than ${\displaystyle 1.\therefore }$ given conic section is hyperbola.
for x-intercept ${\displaystyle \theta =0,\pi }$
when ${\displaystyle \theta =0,r=\frac{5}{2}}$, so x -intercept in this case is ${\displaystyle (\frac{5}{2},0)}$
when ${\displaystyle \theta =\pi .r=\frac{5}{2}}$,so x -intercept in this case is ${\displaystyle (-\frac{5}{2},0)}$
for y-intercept ${\displaystyle \theta =\frac{\pi }{2},\frac{3\pi }{2}}$
when ${\displaystyle \theta =\frac{\pi }{2},r=\frac{5}{7}}$, so y-intercept in this case is ${\displaystyle (0,\frac{5}{7})}$
when

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