Dear Tom,
Nice to hear from you. Don’t worry about the problem. Let me help you solve it step by step.
Since point A(2,1) is on the hyperbola, we put it into
\frac{x^2}{a^2}-\frac{y^2}{a^2-1}=1.
Then we get
\frac4{a^2}-\frac1{a^2-1}=1,
so a^2=2. The hyperbola is
\frac{x^2}{2}-y^2=1.
Suppose the slope of AP is k, then the slope of AQ is -k. By calculating the two intersection points with the hyperbola, we can find that the slope of line l is always
-1.
Next, because
\tan \angle PAQ=2\sqrt2,
we have
\frac{2k}{k^2-1}=2\sqrt2.
So k=\sqrt2. Finally, the area of triangle PAQ is
\frac{16\sqrt2}{9}.
I hope this can help you understand the problem better.
Yours,
Li Lei