Wanna ask 2 polar coordinates Questions

1) The hyperbolic spiral has equation r*(degree)=a for (degree)>0 where a is a positive constant. Using the fact that lim(degree->0 )---(sin(degree)/(de gree))=1 , show that the line y = a is a horizontal asymptote to the spiral. Sketch the spiral.
This is a standard method)
Since x=rcosθ y=rsinθ
sub rθ=a
x=acosθ/θ; y=asinθ/θ
Let θ tends to 0
x=infinity, y=a
So the line y = a is a horizontal asymptote to the spiral
The other method
Since rθ=a
r=a/θ
If we want to find the horizontal asymptote, then the equation is
r=lim(θ->0 )(a/θ)
[Plwase remember the how to find the horizontal asymptote in Cartesian cooedinate]
rsinθ=lim(θ->0 )(asinθ/θ)
rsinθ=a [lim(θ->0 )(sinθ/θ)=1]
y=a [y=rsinθ]
So the line y = a is a horizontal asymptote to the spiral
Sketch the spiral
when a =2, the graph will be

圖片參考：http://upload.wikimedia.org/wikipedia/commons/thumb/b/bf/Mc_HyperbolicSpiral.png/300px-Mc_HyperbolicSpiral.png

2) show that r = 5/(3-2*cos(degree)) is the polar equation of an ellipse by finding the Cartesian equation of the curve( and completing the square).
x=rcosθ y=rsinθ
So
√(x^2+y^2)=5/[3-(2x/√(x^2+y^2)]

√(x^2+y^2)[3-(2x/√(x^2+y^2)]=5
3√(x^2+y^2)-2x=5
3√(x^2+y^2)=5+2x
9(x^2+y^2)=25+20x+4x^2
5x^2+9y^2-20x-25=0
5(x-2)^2+9y^2=45
(x-2)^2/9+y^2/5=1
show that r = 5/(3-2*cos(degree)) is the polar equation of an ellipse. The center is (2,0)


2007-06-09 22:45:36 補充：
plwase 應是 please

2007-06-10 02:46:50 補充：
第一個方法真是不太易﹐我都是找張圖時看到其他人這麼做﹐所以想便抄下來供大家參考。

2007-06-10 02:47:04 補充：
順便

2)
x = r cos@
y = r sin@

r = 5 / (3 - 2cos@)
3r - 2r cos@ = 5
3r - 2x = 5
3r = 2x + 5
9r^2 = 4x^2 + 20x + 25
9(x^2 + y^2) = 4x^2 + 20x + 25
5x^2 + 9y^2 - 20x - 25 = 0
which is ellipse

1)
未諗到,食完飯再諗