91Ó°ÊÓ

The polar equation is

$r=\frac{8}{4+3\sin \left(\mathrm{θ}\right)}$

The polar equation is

$r=\frac{8}{4+3\sin \left(\mathrm{θ}\right)}$

Divide both sides by $4$

$$\begin{gathered} r=\frac{{\displaystyle \frac{8}{4}}}{{\displaystyle \frac{4}{4}}+{\displaystyle \frac{3}{4}}\sin \left(\mathrm{θ}\right)} \\ r=\frac{2}{1+{\displaystyle \frac{3}{4}}\sin \left(\mathrm{θ}\right)} \end{gathered}$$

Compare with the standard polar equation

$r=\frac{ep}{1+e\sin \left(\mathrm{θ}\right)}$

So,

$$\begin{gathered} ep=2 \\ e=\frac{3}{4} \\ \frac{3}{4}×p=2 \\ p=2×\frac{4}{3} \\ p=\frac{8}{3} \end{gathered}$$

Identification of equation:

We have got

$e=\frac{3}{4}<1$

So, this is an ellipse equation.

Finding Directrix:

Apply the directrix formula: $y=p$

Plug $p=\frac{8}{3}$

So, the directrix equation is $y=\frac{8}{3}$

One focus is at the pole, and the directrix is parallel to the polar axis, a distance of $\frac{8}{3}$units to the above pole.

To find the vertices, we let $\mathrm{θ}=0,\mathrm{θ}=\mathrm{π}$and find r

$$\begin{gathered} r=\frac{8}{4+3\sin \left({\displaystyle \frac{\mathrm{π}}{2}}\right)}=\frac{8}{4+3×1} \\ r=\frac{8}{4+3}=\frac{8}{7} \\ r=\frac{8}{4+3\sin \left(\frac{3\mathrm{π}}{2}\right)}=\frac{8}{4+3×-1} \\ r=\frac{8}{4-3}=8 \end{gathered}$$

So, vertices are $(\frac{8}{7},\frac{\mathrm{Ï€}}{2}),(8,\frac{3\mathrm{Ï€}}{2})$

Rectangular forms of vertices are $(0,\frac{8}{7}),(0,-8)$

Now, we can find a midpoint to get center

So, the center is $(0,-\frac{24}{7})$

Now, we can find a and b

a is the distance from center to vertex

So, $a=3$

Now, we can find b

$b=\frac{32}{7}$

Locate vertices $(\frac{8}{7},\frac{\mathrm{Ï€}}{2}),(8,\frac{3\mathrm{Ï€}}{2})$,

draw directrix $y=\frac{8}{3}$

Locate center $(0,-\frac{24}{7})$

Locate distances from the center to vertices$a=3,b=\frac{24}{7}$

So, the graph is