Chapter 10.9, Problem 44E

To determine

To calculate: The polar equation of the parabola whose vertex is at $\left(10,\frac{\pi }{2}\right)$ and focus at the pole.

Answer to Problem 44E

The polar equation of the parabola is $r=\frac{20}{1+\cos \theta }$ .

Explanation of Solution

Given information:

Characteristics ofparabola are vertex is at $\left(10,\frac{\pi }{2}\right)$ and focus at the pole.

Formula used:

When polar equations are either of the form $r=\frac{ep}{1\pm e\sin \theta }$ or $r=\frac{ep}{1\pm e\cos \theta }$ then it is called a conic. Here $e>0$ is the eccentricity and $\left|p\right|$ is the distance between the focus also called pole and the directrix.

When $0<e<1$ , conic is called ellipse.

When $e=1$ , conic is called parabola.

When $e>1$ , conic is called hyperbola.

Also, for the conic $r=\frac{ep}{1+e\cos \theta }$ , vertical directrix is to the right of the pole.

Calculation:

Consider the parabola whose vertex is at $\left(10,\frac{\pi }{2}\right)$ and focus at the pole.

Vertex is the midpoint between directrix and focus. Since, $\left(10,\frac{\pi }{2}\right)$ so the parabola opens rightward. In polar coordinate system origin is referred as pole, so the equation of directrixis $x=20$ . This can be observed from the graph. Blue line is directrix and black dot is vertex.

  

Recall that eccentricity of parabola is $e=1$ .

It is found that equation of directrix is $x=20$ which is a vertical line and lie on right of the pole.

Recall that for the conic $r=\frac{ep}{1+e\cos \theta }$ , vertical directrix is to the right of the pole.

Also it is provided that eccentricity is $1$ and $\left|p\right|$ which is the distance between the focus also called polethe directrix is 20.

Substitute $e=1$ and $p=20$ in the above equation,

  $\begin{array}{c}r=\frac{ep}{1+e\cos \theta }\\ =\frac{1\cdot 20}{1+1\cdot \cos \theta }\\ =\frac{20}{1+\cos \theta }\end{array}$

Thu, the polar equation of the parabola is $r=\frac{20}{1+\cos \theta }$ .