Answer: The polar equation is \(r^2=a^2\cos2\theta\), so the curve is a two-loop lemniscate along the initial line.

The area of one loop is \(\displaystyle \frac{a^2}{2}\).

The tangent is parallel to the initial line when \(\theta=\pm\frac{\pi}{6}\) on the right-hand loop and \(\theta=\pm\frac{5\pi}{6}\) on the left-hand loop, with \(r=\frac{a}{\sqrt2}\).

From

\((x^2+y^2)^2=a^2(x^2-y^2),\)

use \(x=r\cos\theta\), \(y=r\sin\theta\). Then

\(x^2+y^2=r^2,\qquad x^2-y^2=r^2(\cos^2\theta-\sin^2\theta)=r^2\cos2\theta.\)

Hence

\(r^4=a^2r^2\cos2\theta.\)

Apart from the pole,

\(r^2=a^2\cos2\theta.\)

Real points require \(\cos2\theta\geq0\). This gives two loops, one about the initial line and one about the opposite direction.

The right-hand loop is traced for \(-\frac\pi4\leq\theta\leq\frac\pi4\). Its area is

\(A=\frac12\int_{-\pi/4}^{\pi/4}r^2\,\mathrm d\theta=\frac{a^2}{2}\int_{-\pi/4}^{\pi/4}\cos2\theta\,\mathrm d\theta.\)

Thus

\(A=\frac{a^2}{2}\left[\frac12\sin2\theta\right]_{-\pi/4}^{\pi/4}=\frac{a^2}{2}.\)

For a polar curve, a horizontal tangent occurs when

\(\frac{\mathrm dy}{\mathrm d\theta}=r'\sin\theta+r\cos\theta=0.\)

Differentiate \(r^2=a^2\cos2\theta\):

\(2rr'=-2a^2\sin2\theta,\qquad rr'=-a^2\sin2\theta.\)

Multiplying the tangent condition by \(r\) gives

\(rr'\sin\theta+r^2\cos\theta=0.\)

Substitute \(rr'=-a^2\sin2\theta\) and \(r^2=a^2\cos2\theta\):

\(-a^2\sin2\theta\sin\theta+a^2\cos2\theta\cos\theta=0.\)

Using \(\cos2\theta\cos\theta-\sin2\theta\sin\theta=\cos3\theta\),

\(\cos3\theta=0.\)

Hence \(3\theta=\frac\pi2,\frac{3\pi}{2},\ldots\). On the curve this gives

\(\theta=\pm\frac\pi6,\quad \pm\frac{5\pi}{6}.\)

At these values, \(\cos2\theta=\frac12\), so

\(r=\frac{a}{\sqrt2}.\)