Need help with this...thanks in advance...

[Guest]

Show using the following equation:
r = letter l/1 + e*cos theta

a)the polar equations for the circle e=0
b)the parabola, e=1
c)the ellipse (0<e<1)
d)the hyperbola e>1

 

[tkhunny]

Unable to translate...

 

[soroban]

Hello, americo74!

Show using the following equation: \(\displaystyle r\:=\:\frac{de}{1\,+\,e\cdot\cos\theta}\)

a) the polar equation for the circle: \(\displaystyle e\,=\,0\)
b) the parabola: \(\displaystyle e\,=\,1\)
c) the ellipse: \(\displaystyle 0\,<\,e\,<\,1\)
d) the hyperbola: \(\displaystyle e\,>\,1\)

Are we expected to derive the polar equations for all the conics? **

These derivations are given somewhere in your book.
No one . . . no one! . . . is expected to derive these equations "from scratch".
\(\displaystyle \;\;\)(Well, okay ... maybe a truly dedicated math major . . . )


**\(\displaystyle \;\;\) Are we allowed to switch to rectangular coordinates?
\(\displaystyle \;\;\;\;\;\)"Show" is a very vague instruction.
\(\displaystyle \;\;\;\;\;\)Can we demonstrate them with graphs?
\(\displaystyle \;\;\;\;\;\)More information would be welcome.

 

[Guest]

Reply to questions

I thought that the question is meant to be taken literally. That is, show would mean by algebraic methods, don't you agree?

Yes we are expected to derive the polar equations for all the conics...

 

[soroban]

Re: Reply to questions

Hello, americo74!

. . . That is, show would mean by algebraic methods, don't you agree?

Yes, I agree . . . and I'd do it in the simplest way possible.
I would start in rectangular coordinates, then convert to polars.

      :       |       P
     D* - - - + - - - *(x,y)
      :       |     /
      :       |   /
      :       | /
     -+-------*------------
      :       |O
    x=-d

We have the focus at \(\displaystyle O(0,\,0)\) and directrix \(\displaystyle x\,=\,-d\).

Let \(\displaystyle P(x,\,y)\) be any point on the plane such that:
\(\displaystyle \;\;\;\overline{PO}\:=\;e\cdot\overline{PD}\,\) for some real number, \(\displaystyle e\).

Then we have: \(\displaystyle \,\sqrt{x^2\,+\,y^2} \:= \:e(x\,+\,d)\)

In polar coordinates: \(\displaystyle \,r\:=\:e(r\cdot\cos\theta\,+\,d)\;\;\Rightarrow\;\;r\:=\:er\cdot\cos\theta\,+\,de\)

Then: \(\displaystyle \,r\,-\,er\cdot\cos\theta\:=\:de\;\;\Rightarrow\;\;r(1\,-\,e\cdot\cos\theta)\:=\:de\)

Therefore: \(\displaystyle \,r\:=\:\L\frac{de}{1\,-\,e\cdot\cos\theta}\)


(Well, something like that . . . )