r = 1/(3-4sinθ)
(a) see
http://www.wolframalpha.com/input/?i=r+%3D+1%2F%283-4sin%CE%B8%29
(b)
r = 1/(3-4sinθ)
r(3-4sinθ) = 1
3r - 4rsinθ = 1
3√(x^2+y^2) - 4y = 1
3√(x^2+y^2) = 4y+1
3(x^2+y^2) = 16y^2+8y+1
3x^2 - 13y^2 - 8y = 1
3x^2 - 13(y^2 - 8/13 y) = 1
3x^2 - 13(y - 4/13)^2 = 1 + 13(4/13)^2
3x^2 - 13(y - 4/13)^2 = 29/13
39/29 x^2 - 169/29 (y - 4/13)^2 = 1
x^2/(29/39) - (y-4/13)^2/(29/169) = 1
hyperbola with center at (0,4/13)
For the given equation r=1/(3-4sintheta)
a. Graph the equation in polar form.
b. Reduce the equation to simplest rectangular form.
c. Determine the Center and the type of graph of the resulting equation.
3 answers
Steve, your graph does not agree with your solution, so looking at it carefully, I had the same up to
3(√(x^2 + y^2) = 1 + 4y
square both sides
9(x^2 + y^2) = 1 + 8y + 16y^2
9x^2 + 9y^2 - 16y^2 - 8y = 1
9x^2 - 7y^2 - 8y = 1
you forgot to square the 3
I am sure Justine can finish the "completing the square" procedure.
3(√(x^2 + y^2) = 1 + 4y
square both sides
9(x^2 + y^2) = 1 + 8y + 16y^2
9x^2 + 9y^2 - 16y^2 - 8y = 1
9x^2 - 7y^2 - 8y = 1
you forgot to square the 3
I am sure Justine can finish the "completing the square" procedure.
Good catch. I felt vaguely uneasy with the way things worked out, but was in a hurry.
No doubt Justine caught the slip on her own...
No doubt Justine caught the slip on her own...
1 answer