Question

identify the curve by finding a Cartesian equation for the curve.
r^2 cos 2theta = 1

so i'm guessing i use x^2 +y^2 = r^2

and r^2 = 1/cos2theta or sec 2theta

how do i continue?

Answer 1

Use the right triangle reference.

Answer 2

yeah i thought of that
where x^2 = r^2 cos^2 theta
and y^2 = r^2 sin^2 theta
but the 2theta kinda screws up the hope for reduction..unless i use a trig identity
sec

Answer 3

I would use trig identity for that

Answer 4

x^2 = cos^2 theta/ cos 2 theta = 1/1-tan^2 theta (after some simplifying)
and then
y^2 = 1/tan^2 theta - 1

so..
y^2 = -x^2
so x^2 + y^2 = 0

Answer 5

i think that might be wrong actually, solutions manual says its supposed to be a hyperbola but isn't in hyperbola form ?.

Answer 6

o

Answer 7

identity *

Answer 8

i'm trying to see how i would get that in a hyperbola form
(question asks to identify the curve by converting to cartesian )

Answer 9

Oh i seee!

Answer 10

r^2cos2theta=1

So do a trig identity to cos 2 theta
\(cos2\theta=cos^2\theta-sin^2\theta\) plug that in
\(r^2(cos^2\theta-sin^2\theta)=1 \)
No distribute the r^2 in
\(r^2cos^2\theta-r^2sin^2\theta=1 \)
We know that the trig identity for \(rcos\theta=x\) and \(rsin\theta=y\)

Answer 11

omg how did i not see that
zale u genius

Answer 12

\(r^2cos^2\theta-r^2sin^2\theta=1\)
\(x^2=(rcos\theta)^2\)
\(y^2=(rsin\theta)^2\)

Answer 13

ya i got it thnx

Answer 14

No problem.