Answer:
The polar equation is \(r^{2}=a^{2}\sin 2\theta\).
The curve consists of two loops, one in the first quadrant and one in the third quadrant, symmetric about the lines \(\theta=\frac{\pi}{4}\) and \(\theta=-\frac{3\pi}{4}\).
The area enclosed by one loop is \(\frac{1}{2}a^{2}\).
Starting from \((x^{2}+y^{2})^{2}=2a^{2}xy\), use polar coordinates.
Since \(x=r\cos\theta\), \(y=r\sin\theta\), and \(x^{2}+y^{2}=r^{2}\), the equation becomes
\(r^{4}=2a^{2}(r\cos\theta)(r\sin\theta)=2a^{2}r^{2}\sin\theta\cos\theta\).
Using \(2\sin\theta\cos\theta=\sin 2\theta\), this simplifies to
\(r^{4}=a^{2}r^{2}\sin 2\theta\).
For points on the curve with \(r\neq 0\), divide by \(r^{2}\) to get
\(r^{2}=a^{2}\sin 2\theta\).
So the polar equation is \(r^{2}=a^{2}\sin 2\theta\).
To sketch the curve, note that \(r^{2}\ge 0\), so we need \(\sin 2\theta\ge 0\). This happens when \(\theta\) is in \([0,\tfrac{\pi}{2}]\) and in \([-\pi,-\tfrac{\pi}{2}]\). Hence there are two loops: one in the first quadrant and one in the third quadrant. The maximum value of \(r^{2}\) is \(a^{2}\), so the maximum radius is \(a\), occurring when \(\sin 2\theta=1\), i.e. at \(\theta=\frac{\pi}{4}\) and \(\theta=-\frac{3\pi}{4}\).
For the area of one loop, use the polar area formula:
\(\text{Area}=\frac{1}{2}\int r^{2}\,d\theta\).
For the loop in the first quadrant, \(\theta\) runs from \(0\) to \(\frac{\pi}{2}\), so
\(\text{Area}=\frac{1}{2}\int_{0}^{\pi/2} a^{2}\sin 2\theta\,d\theta\).
Now integrate:
\(\frac{1}{2}a^{2}\int_{0}^{\pi/2}\sin 2\theta\,d\theta=\frac{1}{2}a^{2}\left[-\frac{1}{2}\cos 2\theta\right]_{0}^{\pi/2}\).
So
\(\text{Area}=\frac{1}{2}a^{2}\left(-\frac{1}{2}(\cos\pi-\cos 0)\right)=\frac{1}{2}a^{2}\left(-\frac{1}{2}(-1-1)\right)=\frac{1}{2}a^{2}.\)
