Chapter 8.5, Problem 50E

Solution

To find: The correct option from the given options.

The correct option is (A).

Calculation:

The general form of a polar equation is $r=\frac{ke}{1\pm e\sin \theta }$ or $r=\frac{ke}{1\pm e\cos \theta }$ .

On comparing the general form of the polar equation with the given options, it is evident that

  $r=\frac{1}{1+\sin \theta }$ is an equation of a parabola as its eccentricity is $1$ .

  $r=3$ is the polar equation of a circle.

Divide both numerator and denominator of $r=\frac{1}{2-\cos \theta }$ , to bring it to the general form. $r=\frac{0.5}{1-0.5\cos \theta }$ . It is evident that $r=\frac{1}{2-\cos \theta }$ is an equation of ellipse as the eccentricity is less than $1$ .

It is evident that $r=\frac{1}{1+2\cos \theta }$ is an equation of hyperbola, as the eccentricity is greater than $1$ .

Therefore, as $r=1+2\cos \theta$ is not of the general form and it cannot be reduced to the general form, it is not a polar equation of a conic.

Hence, the correct option is A.