Chapter 10, Problem 33RE

To determine

To find:The equivalent polar equation of the Cartesian equation: $x^2+4y^2=16$ .

Answer to Problem 33RE

The equivalent polar equation of the Cartesian equation: $x^2+4y^2=16$ is $r=\frac{4}{\sqrt{1+3{\sin }^2\theta }}$ .

Explanation of Solution

Given information:

The Cartesian equation of the curve is $x^2+4y^2=16$ .

Formula used:

In the Polar co-ordinate system, points are identified by $\left(r,\theta \right)$ where $r$ is the distance of the point to the origin and $\theta$ is the directional angle or the angle from the initial ray which can be considered as the positive $x$ -axis.

Polar-Rectangular Conversion formula:

  $x=r\cos \theta$

  $y=r\sin \theta$

  $r=\sqrt{x^2+y^2}$

  $\tan \theta =\frac{y}{x}$

Calculation:

Let $\left(r,\theta \right)$ be the equivalent polar coordinates of the point $\left(x,y\right)$ where $r$ is the distance of the point to the origin and $\theta$ is the directional angle or the angle from the initial ray which can be considered as the positive $x$ -axis.

Apply the Polar-Rectangular conversion formula and substitute $x=r\cos \theta$ , $y=r\sin \theta$ in the given Cartesian equation $x^2+4y^2=16$ to obtain the equivalent polar equation as:

  $\begin{array}{c}{\left(r\cos \theta \right)}^2+4{\left(r\sin \theta \right)}^2=16\\ r^2{\cos }^2\theta +4r^2{\sin }^2\theta =16\\ r^2{\cos }^2\theta +r^2{\sin }^2\theta +3r^2{\sin }^2\theta =16\\ r^2\left({\cos }^2\theta +{\sin }^2\theta \right)+3r^2{\sin }^2\theta =16\end{array}$

Simplify further to obtain:

  $\begin{array}{c}{r}^2\cdot 1+3r^2{\sin }^2\theta =16\\ r^2+3r^2{\sin }^2\theta =16\\ r^2\left(1+3{\sin }^2\theta \right)=16\\ r^2=\frac{16}{1+3{\sin }^2\theta }\end{array}$

Simplify further:

  $\begin{array}{l}r=\sqrt{\frac{16}{1+3{\sin }^2\theta }}\\ r=\pm \frac{4}{\sqrt{1+3{\sin }^2\theta }}\end{array}$

Since, $r$ is the distance of the point to the origin, it cannot be negative.

  $r=\frac{4}{\sqrt{1+3{\sin }^2\theta }}$

Thus, the equivalent polar equation of the Cartesian equation: $x^2+4y^2=16$ is $r=\frac{4}{\sqrt{1+3{\sin }^2\theta }}$ .