   Chapter 10.6, Problem 43E

Chapter
Section
Textbook Problem

# Finding a Polar Equation In Exercises 39-44. Find a polar equation for the conic with its focus at the pole and the given vertex or vertices.Conic Vertex or VerticesHyperbola ( 1 , 3 π 2 ) ,         ( 9 , 3 π 2 )

To determine

To calculate: The polar form of the equation for the hyperbola having its focus at the pole and the vertices are (1,3π2) and (9,3π2).

Explanation

Given:

The conic is hyperbola having its vertices are (1,3π2) and (9,3π2).

Formula used:

The polar form of the equation for the hyperbola having its focus at the pole is:

r=ed1esinθ

Here e is the eccentricity and d is the distance from the focus to its directrix.

Calculation:

In a hyperbola, eccentricity is greater than 1.

So, e>1

Since vertical is the major axis and the vertices of the hyperbola are (1,3π2) and (9,3π2).

So, the length of major axis would be:

2c=1+9=10

And the length of the minor axis would be:

2a=91=8

Then, the eccentricity of hyperbola is:

e=ca=54

Now, put the value of e in the

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