There's a nice discussion of this topic at
http://laurashears.info/math122/unit4/polarAndParamFormsOfParabola/
change the polar equation r=5/1+cos(theta) to rectangular form.
How do I do this? Thank you! :)
4 answers
recall:
sinØ = y/r
cosØ = x/r
r^2 = x^2 + y^2
I will assume you mean:
r=5/(1+cos(theta) )
r=5/(1+x/r)
r + x = 5
√(x^2 + y^2) = 5-x
confirmation:
http://www.wolframalpha.com/input/?i=%E2%88%9A%28x%5E2+%2B+y%5E2%29+%3D+5-x
http://www.wolframalpha.com/input/?i=+polar+plot+r%3D5%2F%281%2Bcos%28theta%29+%29+from+-4%CF%80+to+4%CF%80
sinØ = y/r
cosØ = x/r
r^2 = x^2 + y^2
I will assume you mean:
r=5/(1+cos(theta) )
r=5/(1+x/r)
r + x = 5
√(x^2 + y^2) = 5-x
confirmation:
http://www.wolframalpha.com/input/?i=%E2%88%9A%28x%5E2+%2B+y%5E2%29+%3D+5-x
http://www.wolframalpha.com/input/?i=+polar+plot+r%3D5%2F%281%2Bcos%28theta%29+%29+from+-4%CF%80+to+4%CF%80
x^2 + y^2 = (5 - x) square right side since x^2 + y^2 = r^2
x^2 + y^2 = 25 - 10x + x^2 the x^2 cancel out
-x^2 -x^2'
y^2 = 25 - 10x is the answer
x^2 + y^2 = 25 - 10x + x^2 the x^2 cancel out
-x^2 -x^2'
y^2 = 25 - 10x is the answer
r=5/1+cos theta