Question

Which conic section is represented by the polar equation r equals Start Fraction 3 over 4 plus 2 sine theta End Fraction?
(1 point)
Responses

ellipse

ellipse

hyperbola

hyperbola

parabola

parabola

circle

Answer 1

To determine which conic section is represented by the polar equation \( r = \frac{3}{4 + 2 \sin \theta} \), we can rewrite this equation in a more recognizable form.

A polar equation of the form \( r = \frac{ed}{1 + e \sin \theta} \) represents a conic section, where \( e \) is the eccentricity, and \( d \) is a constant.

In your equation:
- \( r = \frac{3}{4 + 2 \sin \theta} \)

We can rewrite it to match the standard form:

\[
r(4 + 2 \sin \theta) = 3
\]
\[
4r + 2r \sin \theta = 3
\]

This shows a linear combination of \( r \) and \( r \sin \theta \), which indicates that it is a conic section.

To identify the type of conic section, we can rearrange it in terms of eccentricity:

1. Compare it to the standard form \( r = \frac{ed}{1 - e \sin \theta} \).
- Here, we can identify that the denominator component is \( 4 + 2 \sin \theta\).
- This can be rewritten as \( 4 + 2 \sin \theta = 4\left(1 + \frac{1}{2} \sin \theta\right) \).

2. Now, we can express \( r \) as:
\[
r = \frac{3/4}{1 + \frac{1}{2} \sin \theta}
\]

From this, we can compare it to the standard form:
- Here \( e = \frac{1}{2} \) and \( d = \frac{3}{4} \).

Now, since the eccentricity \( e = \frac{1}{2} < 1 \), this indicates that the conic section is an **ellipse**.

**Final Answer: ellipse**