Question

How do you find the vertex and directrix in the equation 4/(1-sinθ)

Answer 1

you seem to have a polar setup, recall that y = r sin(theta)

Answer 2

I found that the directrix is y=-4 because of the equation

Answer 3

im not familiar with the properties like that :)

r = 4/(1-sin)

r(1-sin) = 4

r-r sin = 4

r - y = 4

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

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

x^2 + y^2 = 16+8y + y^2

x^2 = 16+8y

x^2 - 16 = 8y

(x^2 - 16)8 = y

Answer 4

missed the / when i divide by 8 :/

Answer 5

so it looks to me like an upward parabola, symmetric about the y axis, but shifted down by 16

Answer 6

do you think that I could find the vertex by plotting points?

Answer 7

It has to be polar

Answer 8

you could, but as is it looks to me from my manipulations that when x=0 y=-2

Answer 9

so when r = sqrt(0^2+2^2) = 2

4/(1-sin) = 2

2 = 1-sin

sin = -1 when theta = -pi/2

Answer 10

ok thank you :)

Answer 11

youre welcome ... with any luck those random synapse firings make a little sense :)