Answer
\[
k(\theta) \propto r
\]
Work Step by Step
For $\sqrt{a \cos 2 \theta}=r$
\[
\begin{aligned}
r^{\prime}(\theta)=\frac{d r}{d \theta}=\frac{1}{2 \sqrt{a \cos 2 \theta}}(-2 a \sin 2 \theta) &=-\sqrt{a} \frac{\sin 2 \theta}{\sqrt{\cos 2 \theta}} \\
r^{\prime \prime}(\theta) &=\frac{d^{2} r}{d \theta^{2}}=-\sqrt{a} \frac{\left(\cos ^{2}(2 \theta+1)\right)}{\cos ^{3 / 2}(2 \theta)}
\end{aligned}
\]
$\operatorname{since} \frac{\left|r^{2}+2(d r / d \theta)^{2}-r d^{2} r / d \theta^{2}\right|}{\left(r^{2}+(d r / d \theta)^{2}\right)}=k(\theta)$
\[
\begin{array}{c}
\frac{\left|2 a \frac{\sin ^{2}(2 \theta)}{\cos (2 \theta)}+a \cos (2 \theta)+\sqrt{a} \sqrt{a \cos (2 \theta)} \frac{\cos ^{2}(2 \theta+1)}{\cos ^{3 / 2}(2 \theta)}\right|}{\left|a \cos (2 \theta)+a \frac{\sin ^{2}(2 \theta)}{\cos (2 \theta)}\right|^{3 / 2}} =k(\theta) \\
\left|\frac{\left|2 a\left[\cos ^{2}(2 \theta)+\sin ^{2}(2 \theta)\right]+a\right|\left|\cos (2 \theta)^{3 / 2-1}\right|}{|a|}\right| \cos ^{1 / 2}(2 \theta)= \sqrt{a \cos (2 \theta)}.\frac{3}{\sqrt{a}}=\frac{3}{\sqrt{a}} r(\theta)
\end{array}
\]
So:
\[
k(\theta) \propto r
\]
