Respuesta :
We have the following equation:
[tex]r= \frac{5}{3+2sin(\theta)}[/tex]
If we graph this equation we realize that in fact this is an ellipse with major axis matching the y-axis. So we can recognize these characteristics:
1. Center of the ellipse:
The midpoint C of the line segment joining the foci is called the center of the ellipse. So in this exercise this point is as follows:
[tex]C(0, -2)[/tex]
2. Length of major axis:
The line through the foci is called the major axis, so in the figure if you go from -5, at the y-coordinate, and walk through this major axis to the coordinate 1, the distance you run is the length of the major axis, that is:
[tex]6 \ units[/tex]
3. Length of minor axis:
The line perpendicular to the foci through the center is called the minor axis. So in the figure if you go from -2, at the x-coordinate, and walk through this minor axis to the coordinate 2, the distance you run is the length of the minor axis, that is:
[tex]4 \ units[/tex]
4. Foci:
Let's find c as follows:
[tex]c=\sqrt{a^{2}-b^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}[/tex]
Then the foci are:
[tex]f_{1}=(0, \sqrt{5}-2)[/tex]
[tex]f_{2}=(0, -\sqrt{5}-2)[/tex]
[tex]r= \frac{5}{3+2sin(\theta)}[/tex]
If we graph this equation we realize that in fact this is an ellipse with major axis matching the y-axis. So we can recognize these characteristics:
1. Center of the ellipse:
The midpoint C of the line segment joining the foci is called the center of the ellipse. So in this exercise this point is as follows:
[tex]C(0, -2)[/tex]
2. Length of major axis:
The line through the foci is called the major axis, so in the figure if you go from -5, at the y-coordinate, and walk through this major axis to the coordinate 1, the distance you run is the length of the major axis, that is:
[tex]6 \ units[/tex]
3. Length of minor axis:
The line perpendicular to the foci through the center is called the minor axis. So in the figure if you go from -2, at the x-coordinate, and walk through this minor axis to the coordinate 2, the distance you run is the length of the minor axis, that is:
[tex]4 \ units[/tex]
4. Foci:
Let's find c as follows:
[tex]c=\sqrt{a^{2}-b^{2}}=\sqrt{3^{2}-2^{2}}=\sqrt{5}[/tex]
Then the foci are:
[tex]f_{1}=(0, \sqrt{5}-2)[/tex]
[tex]f_{2}=(0, -\sqrt{5}-2)[/tex]

Answer:
A) ellipse, horizontal directrix at a distance of [tex]\frac{5}{2}[/tex] units above the pole
Step-by-step explanation:
I just took the test.