Conics in polor form
- Thread starter kaebun
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Too bad you have not defined e, k, a, b, or c.kaebun said:
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.
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]
\(\displaystyle 3sqrt(x^2+y^2)-x=4 =
3sqrt(x^2+y^2)=4+x =
9(x^2+y^2)=(4+x)^2 =
9x^2+9y^2=16+8x+x^2 =
8x^2+9y^2-8x=16 =
(x-1)^2/3+y^2/2.67=1\)
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]
Hello, kaebun!
Let's start it over . . .
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 \(\displaystyle \left(\frac{1}{2},\,0\right),\;\;a\,=\,\frac{3}{2},\;\;b\,=\,\sqrt{2}\)
Let's start it over . . .
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
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\(\displaystyle 3\sqrt{x^2 + y^2} - x = 4\)
\(\displaystyle 3\sqrt{x^2 + y^2} = x + 4\)
\(\displaystyle 9(x^2 + y^2) = x^2 + 8x + 16\)
\(\displaystyle 8x^2 + 9y^2 - 8x - 16 = 0\)
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
\(\displaystyle 3\sqrt{x^2 + y^2} = x + 4\)
\(\displaystyle 9(x^2 + y^2) = x^2 + 8x + 16\)
\(\displaystyle 8x^2 + 9y^2 - 8x - 16 = 0\)
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