Conics in polor form

[kaebun]

$\displaystyle r=4/(3-cost)$ rewrite in Cartesian coordinates i figured out that e= 1/3
and k=4 but I'm not sure how to get a, b, and c
please help!

 

[tkhunny]

$\displaystyle r=4/(3-cost)$ rewrite in Cartesian coordinates i figured out that e= 1/3
and k=4 but I'm not sure how to get a, b, and c
please help!

Too bad you have not defined e, k, a, b, or c.

Multiply (3-cos(t)) never is zero (0).

r(3-cos(t)) = 4
3r - r*cos(t) = 4

Substitute x = r*cos(t)
Substitute r = sqrt(x^2 + y^2)

3(sqrt(x^2 + y^2)) - x = 4

See what you can do to make it look more familiar.

 

[kaebun]

ok so i put that in standard form
\(\displaystyle 3sqrt(x^2+y^2)-x=4 =
3sqrt(x^2+y^2)=4+x =
9(x^2+y^2)=(4+x)^2 =
9x^2+9y^2=16+8x+x^2 =
8x^2+9y^2-8x=16 =
(x-1)^2/3+y^2/2.67=1\)
this is not the answer in the back of the book
the original equation is in the form $\displaystyle r=ek/91-ecost$
and I am trying to put itinto the form Ax^2+Bxy+Cy^2+Dx+Ey+F=0 so thats what i ment by finding a,b and c etc
I like your way better, maybe I messed up the algebra?[/quote]

 

[soroban]

Hello, kaebun!

Let's start it over . . .

Write in Cartesian corrdinates: $\displaystyle \,r\:=\:\frac{4}{3\,-\,\cos\theta}$

We have: $\displaystyle \,3r\,-\,r\cdot\cos\theta\;=\;4\;\;\Rightarrow\;\;3r\;=\;r\cdot\cos\theta\,+\,4$

Then: $\displaystyle \,3\sqrt{x^2\,+\,y^2}\;=\;x\,+\,4$

Square: \(\displaystyle \,9(x^2\,+\,y^2}\;=\;x^2\,+\,8x\,+\,16\;\;\Rightarrow\;\;9x^2\,+\,9y^2\;=\;x^2\,+\,8x\,+\,16\)

Hence, we have: \(\displaystyle \L\;8x^2\,+\,9y^2\,-\,8x\,-\,16\;=\;0\)

$\displaystyle \;\;\;A\,=\,8,\;\;B\,=\,0.\;\;C\,=\,9\,\;\;D\,=\,-8,\;\;E\,=\,0,\;\;F\,=\,-16$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

We have: \(\displaystyle \:8x^2\,-\,8x\,+\,9y^2\;=\;16 \;\;\Rightarrow\;\;8(x^2\,-\,x\;\;\\,+\,9y^2\;=\;16\)

Complete the square: $\displaystyle \,8\left(x^2\,-\,x\,+\,\frac{1}{4}\right)\,+\,9y^2 \;= \;16\,+\,2$

Hence: $\displaystyle \:8\left(x\,-\,\frac{1}{2}\right)^2\,+\,9y^2\;=\;18$

Divide by 18: \(\displaystyle \L\:\frac{\left(x\,-\,\frac{1}{2}\right)^2}{\frac{9}{4}}\,+\,\frac{y^2}{2}\;=\;1\)

Therefore: $\displaystyle \;$Center $\displaystyle \left(\frac{1}{2},\,0\right),\;\;a\,=\,\frac{3}{2},\;\;b\,=\,\sqrt{2}$

 

[skeeter]

$\displaystyle 3\sqrt{x^2 + y^2} - x = 4$
$\displaystyle 3\sqrt{x^2 + y^2} = x + 4$
$\displaystyle 9(x^2 + y^2) = x^2 + 8x + 16$
$\displaystyle 8x^2 + 9y^2 - 8x - 16 = 0$
is that what you're looking for?


complete the square ...

$\displaystyle 8(x^2 - x + \frac{1}{4}) + 9y^2 = 16 + 2$
$\displaystyle 8(x - \frac{1}{2})^2 + 9y^2 = 18$
$\displaystyle \frac{4(x - \frac{1}{2})^2}{9} + \frac{y^2}{2} = 1$

and you get the ellipse centered at $\displaystyle (\frac{1}{2}, 0)$ with
$\displaystyle a = \frac{3}{2}$ and $\displaystyle b = \sqrt{2}$

Soroban beat me ... still getting used to the Latex

 

[tkhunny]

ok so i put that in standard form
\(\displaystyle 3sqrt(x^2+y^2)-x=4 =
3sqrt(x^2+y^2)=4+x =
9(x^2+y^2)=(4+x)^2 =
9x^2+9y^2=16+8x+x^2 =
8x^2+9y^2-8x=16 =
(x-1)^2/3+y^2/2.67=1\)

Whatever that means -- never write it again. Yikes!

 

[kaebun]

ohhhhhhhhhhhh i messed up completeing the square apparently 1/2= 1 in my brain i feel dumb now