(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 \(\displaystyle{r}=\frac{{{e}{d}}}{{{1}+{e} \cos{\theta}}}\), we get

\(\displaystyle{e}=\frac{3}{{7}}\), since e i.e. eccentricity lies between 0 & 1. :. given conic section is ellipse.

for x-intercept \(\displaystyle\theta={0}\),

\(\displaystyle\pi\ {w}{h}{e}{n}\ \theta={0},{r}=\frac{5}{{10}}=\frac{1}{{2}},\) 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 \(\displaystyle{\left(-\frac{5}{{4}},{0}\right)}\)

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 \(\displaystyle{\left({0},\frac{5}{{7}}\right)}\)

when \(\displaystyle\theta=\frac{{{3}\pi}}{{2}},{r}=\frac{5}{{7}}\), so y-intercept in this case is \(\displaystyle{\left({0},-\frac{5}{{7}}\right)}\)

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 \(\displaystyle={\left(\frac{{\frac{1}{{2}}-\frac{5}{{4}}}}{{2}},{0}\right)}\)

\(\displaystyle={\left(-\frac{3}{{8}},{0}\right)}\)

Now we will distance of center \(\displaystyle{\left(-\frac{3}{{8}},{0}\right)}\) 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 \(\displaystyle{\left(-\frac{3}{{8}}-\frac{3}{{8}},{0}\right)}\)

i.e. \(\displaystyle{\left(-\frac{6}{{8}},{0}\right)}\)

so, x intercepts are \(\displaystyle{\left(\frac{1}{{2}},{0}\right)},{\left(-\frac{5}{{4}},{0}\right)}\)

y intercepts are \(\displaystyle{\left({0},\frac{5}{{7}}\right)},{\left({0},-\frac{5}{{7}}\right)}\)

foci are \(\displaystyle{\left({0},{0}\right)},{\left(-\frac{6}{{8}},{0}\right)}\)

(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 \(\displaystyle{r}=\frac{{{e}{d}}}{{{1}+{e} \sin{\theta}}}\), 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 \(\displaystyle{\left(\frac{5}{{2}},{0}\right)}\)

when \(\displaystyle\theta=\pi.{r}=\frac{5}{{2}}\),so x -intercept in this case is \(\displaystyle{\left(-\frac{5}{{2}},{0}\right)}\)

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 \(\displaystyle{\left({0},\frac{5}{{7}}\right)}\)

when \(\displaystyle\theta=\frac{{{3}\pi}}{{2}},{r}=-\frac{5}{{3}}\), since r is negative .so y-intercept in this case is \(\displaystyle{\left({0},\frac{5}{{3}}\right)}\)

One foci is (0, 0) of this form. Now, to find other foci we will find out center of the hyperbola & it will be the mid point of both the y-intercepts

i.e. center \(\displaystyle={\left({0},\frac{{\frac{5}{{7}}+\frac{5}{{3}}}}{{2}}\right)}\)

\(\displaystyle={\left({0},\frac{25}{{21}}\right)}\)

Now we will distance of center \(\displaystyle{\left({0},\frac{25}{{21}}\right)}\) and 1st foci \(\displaystyle{\left({0},{0}\right)},{c}=\frac{25}{{21}}\)

since the distance of both the foci from the center is same, so coordinates of 2nd foci is \(\displaystyle{\left({0},\frac{25}{{21}}+\frac{25}{{21}}\right)}{i}.{e}.{\left({0},\frac{50}{{21}}\right)}\)

so, x intercepts are \(\displaystyle{\left(\frac{5}{{2}},{0}\right)},{\left(-\frac{5}{{2}},{0}\right)}\)

y intercepts are \(\displaystyle{\left({0},\frac{5}{{7}}\right)},{\left({0},\frac{5}{{7}}\right)}\)

foci are \(\displaystyle{\left({0},{0}\right)},{\left({0},\frac{50}{{21}}\right)}\)

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 \(\displaystyle{r}=\frac{{{e}{d}}}{{{1}+{e} \cos{\theta}}}\), we get

\(\displaystyle{e}=\frac{3}{{7}}\), since e i.e. eccentricity lies between 0 & 1. :. given conic section is ellipse.

for x-intercept \(\displaystyle\theta={0}\),

\(\displaystyle\pi\ {w}{h}{e}{n}\ \theta={0},{r}=\frac{5}{{10}}=\frac{1}{{2}},\) 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 \(\displaystyle{\left(-\frac{5}{{4}},{0}\right)}\)

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 \(\displaystyle{\left({0},\frac{5}{{7}}\right)}\)

when \(\displaystyle\theta=\frac{{{3}\pi}}{{2}},{r}=\frac{5}{{7}}\), so y-intercept in this case is \(\displaystyle{\left({0},-\frac{5}{{7}}\right)}\)

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 \(\displaystyle={\left(\frac{{\frac{1}{{2}}-\frac{5}{{4}}}}{{2}},{0}\right)}\)

\(\displaystyle={\left(-\frac{3}{{8}},{0}\right)}\)

Now we will distance of center \(\displaystyle{\left(-\frac{3}{{8}},{0}\right)}\) 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 \(\displaystyle{\left(-\frac{3}{{8}}-\frac{3}{{8}},{0}\right)}\)

i.e. \(\displaystyle{\left(-\frac{6}{{8}},{0}\right)}\)

so, x intercepts are \(\displaystyle{\left(\frac{1}{{2}},{0}\right)},{\left(-\frac{5}{{4}},{0}\right)}\)

y intercepts are \(\displaystyle{\left({0},\frac{5}{{7}}\right)},{\left({0},-\frac{5}{{7}}\right)}\)

foci are \(\displaystyle{\left({0},{0}\right)},{\left(-\frac{6}{{8}},{0}\right)}\)

(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 \(\displaystyle{r}=\frac{{{e}{d}}}{{{1}+{e} \sin{\theta}}}\), 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 \(\displaystyle{\left(\frac{5}{{2}},{0}\right)}\)

when \(\displaystyle\theta=\pi.{r}=\frac{5}{{2}}\),so x -intercept in this case is \(\displaystyle{\left(-\frac{5}{{2}},{0}\right)}\)

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 \(\displaystyle{\left({0},\frac{5}{{7}}\right)}\)

when \(\displaystyle\theta=\frac{{{3}\pi}}{{2}},{r}=-\frac{5}{{3}}\), since r is negative .so y-intercept in this case is \(\displaystyle{\left({0},\frac{5}{{3}}\right)}\)

One foci is (0, 0) of this form. Now, to find other foci we will find out center of the hyperbola & it will be the mid point of both the y-intercepts

i.e. center \(\displaystyle={\left({0},\frac{{\frac{5}{{7}}+\frac{5}{{3}}}}{{2}}\right)}\)

\(\displaystyle={\left({0},\frac{25}{{21}}\right)}\)

Now we will distance of center \(\displaystyle{\left({0},\frac{25}{{21}}\right)}\) and 1st foci \(\displaystyle{\left({0},{0}\right)},{c}=\frac{25}{{21}}\)

since the distance of both the foci from the center is same, so coordinates of 2nd foci is \(\displaystyle{\left({0},\frac{25}{{21}}+\frac{25}{{21}}\right)}{i}.{e}.{\left({0},\frac{50}{{21}}\right)}\)

so, x intercepts are \(\displaystyle{\left(\frac{5}{{2}},{0}\right)},{\left(-\frac{5}{{2}},{0}\right)}\)

y intercepts are \(\displaystyle{\left({0},\frac{5}{{7}}\right)},{\left({0},\frac{5}{{7}}\right)}\)

foci are \(\displaystyle{\left({0},{0}\right)},{\left({0},\frac{50}{{21}}\right)}\)