Question

Convert the rectangular equation to a polar equation.

y^2-x^2=7sqrt(x^2+y^2)

Answer 1

\(x = r\cos(\theta)\)
\(y = r\sin(\theta)\)

Substitute and Simplify.

Answer 2

i know that.. but what i got was -rcos(2theta) = 7r
and it says the answer is r=-7/cos(2theta)

Answer 3

Why didn't you say so? If you show your work, we can have a much better conversation much more quickly.

\(x^{2} = r^{2}\cos^{2}(\theta)\)

\(y^{2} = r^{2}\sin^{2}(\theta)\)

\(\sqrt{x^{2} + y^{2}} = r\)

And we're left with: \(r(\sin^{2}(x) - cos^{2}(x)) = 7\)

Sure enough, exctly as you have it. So, what's the dilemma? Not the form allowed by the exam? Try cos(x) = 1/sec(x)