Convert the polar equation to a rectangular equation: r = (8secθ) / (4secθ + 1)?

Question

Convert the polar equation to a rectangular equation.
r = (8secθ) / (4secθ + 1)

Simplify the rectangular equation by moving all of the terms to the left side of the equation, and combining like terms. The right side of the equation will then be 0.

??? = 0


I would like to know how to solve step-by-step to this problem. The answer to this question is:
15x^2 + 16y^2 + 16x - 64 = 0

thanks so much

Answer 1

we have x^2+y^2 = r^2 for polar coordinates'

x = r cos t
y = r sin t

we need to eliminate t ( t for theta as I cannot key in theta)

1st get in terms of cos sin

r = 8/(4+ cos t)

4r + r cos t = 8

but r cos t = x

so 4r = 8- x
square both sides
16r^2 = 64+x^2-16x

using r^2 = x^2 + y^2

16(x^2+y^2) = 64 + x^2 -16x

16x^2+16y^2 = 64 + x^2 - 16x
or
tarnsferring RHS to left

15x^2+16x+16y^2- 64 = 0

Answer 2

Well you know x=r cosθ , y = r sinθ.
We want to eliminate both r and θ.

So use the substitutions:
r = sqrt(x^2+y^2)
cosθ = x/r = x/sqrt(x^2+y^2)
secθ = r/x = sqrt(x^2+y^2)/x

So:
r(4secθ + 1) = (8secθ)
r(4 + cosθ) = 8
sqrt(x^2+y^2) (4 + x/sqrt(x^2+y^2)) = 8
Square both sides:
(x^2+y^2) (16 + 8x/sqrt(x^2+y^2) + x^2/(x^2+y^2)) = 64
Multiply out:
16 (x^2+y^2) + 8x.sqrt(x^2+y^2) + x^2 = 64
17x^2 + 16y^2 + 8x.sqrt(x^2+y^2) = 64

Not quite:
15x^2 + 16y^2 + 16x - 64 = 0

Try 2:
Note also: r^2 = r^2 (cos^2 θ sin^2 θ) = x^2 + y^2
r = (8secθ) / (4secθ + 1)
= 8/ (4 + cosθ)
r(4 + cosθ) = 8
r^2(16 + 8cosθ +cos^2 θ) = 64
16r^2 + 8r^2 cosθ + r^2 cos^2 θ = 64
16(x^2 + y^2) + 8r (r cosθ) + (r cosθ)^2 = 64
16(x^2 + y^2) + 8r (x) + x^2 = 64
That is not successful, still have a factor 8rx (= 8x.sqrt(x^2+y^2)).
Also the extra x^2 term is +1 not -1?

Dupinder has it, the trick was to move the r cosθ term to the RHS to avoid getting the awkward 8x.sqrt(x^2+y^2) cross-term after you square.

Answer 3

r = (8secθ) / (4secθ + 1)
convert sec into cos
r=(8/cosθ)/(4/cosθ+1)
r=8/(4+cosθ)
4r+rcosθ=8
4r+rcosθ-8=0
put r=sq. root of(x^2+y^2)
rcosθ=x
4 sq. root of(x^2+y^2)+x-8=0
4 sq. root of(x^2+y^2)=8-x
squaring both sides u get the ans.