   Chapter 10.6, Problem 55E

Chapter
Section
Textbook Problem

# Finding a Polar Equation In Exercises 51-54, use the results of Exercises 49 and 50 to write the polar form of the equation of the conic. x 2 9 − y 2 16 = 1

To determine

To calculate: The polar form of the equation of conic having equation x29y216=1 using the result of exercise 50. That is the polar form of the equation for a hyperbola is r2=b21e2cos2θ

Explanation

Given:

The equation is x29y216=1.

Formula used:

The rectangular form of the hyperbola is x2a2y2b2=1 where ab has foci (±c,0).

Here c2=a2+b2 eccentricity e=1+b2a2 and vertices (±a,0) and its polar form is r2=b21e2cos2θ.

Calculation:

Consider the equation:

x29y216=1

Now compare this equation with standard equation of hyperbola that is x2a2y2b2=1.

So, the value of a is 3 and b is 4

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