Chapter 10.6, Problem 5AYU

To determine

The choice in the blank for the following statement

If $\left(r,\theta \right)$ are polar coordinates, the equation $r=\frac{2}{2+3\sin \theta }$ defines a (an) $\underset{\_}{}$

Parabola (b) hyperbola (c) ellipse (d) circle

Answer to Problem 5AYU

Solution:

If $\left(r,\theta \right)$ are polar coordinates, the equation $r=\frac{2}{2+3\sin \theta }$ defines a hyperbola.

Explanation of Solution

Given information:

The equation is $r=\frac{2}{2+3\sin \theta }$

Explanation:

Theorem:

The polar equation of conics with focus at the pole and directrix is perpendicular to the polar axis at a distance $p$ units the right of the pole is $r=\frac{ep}{1+e\sin \theta }$, where $e$ is the eccentricity of the conic.

If $e=1$, then conic is a parabola and the axis of symmetry is perpendicular to the directrix.

If $e>1$, then conic is a hyperbola and the axis of symmetry is perpendicular to the directrix.

If $e<1$, then conic is an ellipse and the axis of symmetry is perpendicular to the directrix.

Convert the given equation $r=\frac{2}{2+3\sin \theta }$ into $r=\frac{ep}{1+e\sin \theta }$

Divide numerator and denominator on both sides by $2$

$r=\frac{1}{1+\frac{3}{2}\sin \theta }$

Compare the given equation $r=\frac{1}{1+\frac{3}{2}\sin \theta }$ with $r=\frac{ep}{1+e\sin \theta }$

$\Rightarrow e=\frac{3}{2}=1.5$ and $ep=1$

$\frac{3}{2}\cdot p=1$

By simplifying,

$\Rightarrow p=\frac{2}{3}$.

Thus, $e=\frac{3}{2}=1.5>1$, the conic is a hyperbola with focus at the pole and directrix perpendicular to the polar axis at a distance $p=1$ units the right of the pole.