Find a Cartesian equation for the curve and identify it. $ r^2 \cos 2\theta = 1$
Find a Cartesian equation for the curve and identify it. $$ r^2 \cos 2\theta = 1$$
I'm confused by the $2\theta.$
I isolated $r^2$ to get $r^2 = \frac{1}{\cos2\theta}$
Now, normally if it was just a $\cos \theta$ I would multiply both sides by $\frac{1}{r}$ and then substitute $r$ for $\sqrt {x^2+y^2}$ and then substitute $r\cos\theta$ for $x$. But the $2\theta$ doesn't allow for that to happen.
$\endgroup$ 13 Answers
$\begingroup$Since $\cos 2 \theta = 2 \cos^2 \theta -1$, then $$r^2 \cos 2 \theta = 2 r^2 \cos^2 \theta - r^2 = 2(r \cos \theta)^2 - r^2 = 2x^2 - (x^2 + y^2) = x^2 - y^2 ,$$ so your curve is $x^2 - y^2 = 1$ which is known to be a hyperbola having the lines $y = \pm x$ as asymptotes.
$\endgroup$ $\begingroup$Using $cos2\theta$ = $cos^2\theta - sin^2\theta$,
$r^2(cos^2\theta - sin^2\theta$) =1
$(rcos\theta)^2$ - ($rsin\theta)^2$ = 1
$x^2 - y^2 = 1$.
This is a hyperbola centred at the origin.
$\endgroup$ $\begingroup$There are a few ways to approach, but here's a hint for how I'd do it: $\cos 2\theta = 2\cos^2\theta - 1$, and $x = r \cos\theta$, so $x^2 = \dots$, and so...
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