Chapter 10, Problem 11CT

To determine

The conic represented by the polar equation $r=\frac{3}{1-2\cos \theta }$ and find the rectangular equation.

Answer to Problem 11CT

Solution:

The conic represented by the polar equation $r=\frac{3}{1-2\cos \theta }$hyperbola and the equation of hyperbola in rectangular coordinates is $\frac{{\left(x+2\right)}^2}{1}-\frac{y^2}{3}=1$ .

Explanation of Solution

Given information:

The polar equation, $r=\frac{3}{1-2\cos \theta }$

Theorem:

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

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

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

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

Compare the given equation $r=\frac{3}{1-2\cos \theta }$ with $r=\frac{ep}{1-e\cos \theta }$ ,

  $\Rightarrow e=2$ and $ep=3$

By simplifying,

  $\Rightarrow p=\frac{3}{2}$ .

Therefore, $e=2>1$, the conic is a hyperbola.

Now, $r=\frac{3}{1-2\cos \theta }$

Rearrange the equation,

  $\Rightarrow r(1-2\cos \theta )=3$

Distribute $r$

  $\Rightarrow r-2r\cos \theta =3$

  $\Rightarrow r=3+2r\cos \theta$

Square on both sides,

  $\Rightarrow r^2={\left(3+2r\cos \theta \right)}^2$

Since $x=r\cos \theta$ and $x^2+y^2=r^2$

  $\Rightarrow x^2+y^2={\left(3+2x\right)}^2$

  $\Rightarrow x^2+y^2=9+12x+4x^2$

  $\Rightarrow 3x^2+12x-y^2=-9$

Completing the square in $x$ ,

  $\Rightarrow \left(3x^2+12x+12\right)-y^2=-9+12$

  $\Rightarrow 3\left(x^2+4x+4\right)-y^2=3$

  $\Rightarrow 3{\left(x+2\right)}^2-y^2=3$

Divide by $3$ on both sides,

  $\Rightarrow \frac{{\left(x+2\right)}^2}{1}-\frac{y^2}{3}=1$

Therefore, $\frac{{\left(x+2\right)}^2}{1}-\frac{y^2}{3}=1$ is the equation of hyperbola in rectangular coordinates