Chapter 9: Q. 84 (page 584)

Show that the graph of the equation $r = - 2 a \sin θ$ , ,
is a circle of radius a with center at $(0, - a)$ in rectangular
coordinates.

Short Answer

Expert verified

$x^{2} + {(y + a)}^{2} = a^{2}$

Step by step solution

Step 1. Given

given polar coordinate

$r = - 2 a \sin θ$

Step 2. Concept

To convert given polar coordinate into rectangular coordinate we try to deduce the given polar coordinate into the following expression:

$x^{2} + y^{2} = 2 a y$

Step 3. Calculation

Consider the given polar coordinate

$r = - 2 a \sin θ m u l t i p l y t h e a b o v e e q u a t i o n b y r r^{2} = - 2 a . r \sin θ s u b s t i t u t e x^{2} + y^{2} i n p l a c e o f r^{2} a n d r \sin θ f o r y w e g e t x^{2} + y^{2} = - 2 a y x^{2} + y^{2} + 2 a y = 0 N o w a d d a^{2} b o t h s i d e s t o m a k e i t a p e r f e c t s q u a r e x^{2} + (y^{2} + 2 a y + a^{2}) = a^{2} x^{2} + {(y + a)}^{2} = a^{2} w h i c h i s o f t h e f o r m o f e q u a t i o n o f a c i r c l e .$

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Show that the graph of the equation $r = - 2 a \sin θ$ , ,
is a circle of radius a with center at $(0, - a)$ in rectangular
coordinates.

Short Answer

Step by step solution

Step 1. Given

Step 2. Concept

Step 3. Calculation

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Show that the graph of the equation r=-2asinθ, ,is a circle of radius a with center at (0,-a)in rectangularcoordinates.

Short Answer

Step by step solution

Step 1. Given

Step 2. Concept

Step 3. Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Mechanics Maths

Logic and Functions

Calculus

Discrete Mathematics

Geometry

Decision Maths

Study anywhere. Anytime. Across all devices.

Show that the graph of the equation $r = - 2 a \sin θ$ , ,
is a circle of radius a with center at $(0, - a)$ in rectangular
coordinates.