r times Secant theta equals -5

Question

put into polar form

Answer 1

If you want it in polar than George and Bill are both right. But just incase you need it in RECTANGULAR then:

 

Given: rSec(θ)=-5

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:

y=(-5x-x²)^1/2

That is if you wanted it in RECTANGULAR FORM... but as for polar form it is already in polar form. 

Answer 2

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)

Answer 3

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.