I asked the same question twice on Bartleby and received two different solutions, and I just need to know which one is correct, the original problem is this: Convert the given polar equation into an equation in terms of Cartesian coordinates.r = (6sec(theta))/(−2 + 3sec(theta))And screenshots of the two answers I received are attached. =r =yx* +y°Now use the idetitity sec(0)cos(0)cos(0)3-2+cos(0)cos (0)-2 cos (0) +3cos (0)cos (0)a dUsecos (0)" -2cos(0)+3|-2 cos (0) +3-2 cos(0)+3Step 2Now multiply both sides by (-2 cos(0)+3) and further simplify itr(-2cos(0) + 3)=·(-2cos(0)+3)(-2cos(0)+3)r(-2cos(0)+3)=6Now clear the parenthesis and the multiply-2r cos(0)+3r = 6Substitute rcos (0) =x and r = Vx² +y²-2.x + 3/x² + y° = 6Which is required cartesian form. UIVei polar equauon is r-2+3 sec e)Note that x=rcos 0, y =r sino.Obtain r =.r(-2+3sec 0) = 6 sece-2r + 3r sece =6 sec eЗr-2r +cos e-2r cos 0 +3rcos 0-2x+3r = 6sec ecose-2x+3r = 6Зr 3D6-2хStep 2On further simplification,6sec 0-2+3 sece6-2х)(x+y)* -(36- 24x +4xx* +y° +2xy =9x +9y² +18xy = 36 – 24x + 4x²5x3 +9у? +18ху+ 24х-36%3D0Thus, the Cartesian equation is 5x +9y* +18y+24x- 36 = 0.

Answered: Image /qna-images/answer/864c9fa0-37a0-4879-9733-940f5461e02b.jpg

Answered: =r =yx* +y° Now use the idetitity…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

See similar textbooks

Related questions

Question

100%

I asked the same question twice on Bartleby and received two different solutions, and I just need to know which one is correct, the original problem is this:

Convert the given polar equation into an equation in terms of Cartesian coordinates.

r = (6sec(theta))/(−2 + 3sec(theta))

And screenshots of the two answers I received are attached.

=r =yx* +y°
Now use the idetitity sec(0)
cos(0)
cos(0)
3
-2+
cos(0)
cos (0)
-2 cos (0) +3
cos (0)
cos (0)
a d
Use
cos (0)" -2cos(0)+3|
-2 cos (0) +3
-2 cos(0)+3
Step 2
Now multiply both sides by (-2 cos(0)+3) and further simplify it
r(-2cos(0) + 3)=
·(-2cos(0)+3)
(-2cos(0)+3)
r(-2cos(0)+3)=6
Now clear the parenthesis and the multiply
-2r cos(0)+3r = 6
Substitute rcos (0) =x and r = Vx² +y²
-2.x + 3/x² + y° = 6
Which is required cartesian form.

UIVei polar equauon is r
-2+3 sec e)
Note that x=rcos 0, y =r sino.
Obtain r =.
r(-2+3sec 0) = 6 sece
-2r + 3r sece =6 sec e
Зr
-2r +
cos e
-2r cos 0 +3r
cos 0
-2x+3r = 6sec ecose
-2x+3r = 6
Зr 3D6-2х
Step 2
On further simplification,
6sec 0
-2+3 sece
6-2х)
(x+y)* -(
36- 24x +4x
x* +y° +2xy =
9x +9y² +18xy = 36 – 24x + 4x²
5x3 +9у? +18ху+ 24х-36%3D0
Thus, the Cartesian equation is 5x +9y* +18y+24x- 36 = 0.

Expert Solution

Step by step

Solved in 5 steps with 5 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Polar Equations of Conics$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

Similar questions