put into polar form

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 r^{2}/x=-5 ^{
}

we know by Pythagorean theorem that r^{2}=x^{2}+y^{2}

so we really have (x^{2}+y^{2})/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-x^{2})^{1/2}

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