put into polar form
r times Secant theta equals -5
If you want it in polar than George and Bill are both right. But just incase you need it in RECTANGULAR then:
we know that Cos(θ)=x/r since Sec(θ) =1/Cos(θ) then Sec(θ)=r/x
so plug rSec(θ)= r(r/x)
that means that r2/x=-5
we know by Pythagorean theorem that r2=x2+y2
so we really have (x2+y2)/x=-5
Which is sufficient as for rectangular form. If you wanted the so solve the problem as a function of x only then it would be:
That is if you wanted it in RECTANGULAR FORM... but as for polar form it is already in polar form.
Hi Jen... looks to be like this already is in polar (nor rectangular coordinate) form; just need to rearrange to solve for r:
r * Sec(θ) = -5; r = -5/Sec(θ) = -5 / (1/Cos(θ)) = -5Cos(θ)
r sec Θ = -5
-(r/5) = cos Θ
arccos (-r/5) = Θ, Θ lies in the 2nd quadrant and depends upon the value of r.