Answer: (a) Greatest distance from the pole is \(1\).

(b) \(\dfrac18\ln2\)

(c) \(x^4-2xy-y^4=0\)

(d) \(\dfrac18\sqrt2-\dfrac18\ln2\)

(a)

Since \(r^2=\tan2\theta\), at \(\theta=0\), \(r=0\), so the curve starts at the pole. As \(\theta\) increases to \(\pi/8\), \(\tan2\theta\) increases to \(\tan(\pi/4)=1\), so \(r\) increases to \(1\).

The greatest distance from the pole is therefore \(1\).

(b)

For polar area,

\(A=\frac12\int r^2\,d\theta.\)

Here \(r^2=\tan2\theta\), with \(0\le\theta\le\pi/8\), so

\(A=\frac12\int_0^{\pi/8}\tan2\theta\,d\theta.\)

Since \(\int\tan2\theta\,d\theta=-\frac12\ln(\cos2\theta)\),

\(A=\left[-\frac14\ln(\cos2\theta)\right]_0^{\pi/8}.\)

Thus

\(A=-\frac14\ln\left(\frac{\sqrt2}{2}\right)=\frac18\ln2.\)

(c)

Use

\(\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}.\)

Since \(\tan\theta=\frac yx\),

\(r^2=\frac{2(y/x)}{1-y^2/x^2}=\frac{2xy}{x^2-y^2}.\)

Also \(r^2=x^2+y^2\). Therefore

\(x^2+y^2=\frac{2xy}{x^2-y^2}.\)

Multiplying by \(x^2-y^2\),

\((x^2+y^2)(x^2-y^2)=2xy.\)

Hence

\(x^4-y^4=2xy,\)

so

\(x^4-2xy-y^4=0.\)

(d)

The required area is the area of the triangle formed by the pole, the point \(r=1,\theta=\pi/8\), and its projection on the \(x\)-axis, minus the polar area from part (b).

The triangle area is

\(\frac12\cos\frac{\pi}{8}\sin\frac{\pi}{8}=\frac14\sin\frac{\pi}{4}=\frac{\sqrt2}{8}.\)

Therefore the required area is

\(\frac{\sqrt2}{8}-\frac18\ln2.\)

Mark scheme