The point with coordinates \((a, b)\), where \(a\) and \(b\) are rational numbers,is called an integer rational point if both \(a\) and \(b\) are integers; a non-integer rational point if neither \(a\) nor \(b\) is an integer.
\(\bf (a)\) Write down an integer rational point and a non-integer rational point on the circle \(x^2+y^2 =1\).
[\bf (b)] Write down an integer rational point on the circle
\(x^2+y^2=2\).
Simplify
\[
(\cos\theta + \sqrt m \sin\theta)^2 + (\sin\theta - \sqrt m \cos\theta)^2 \,
\]
and hence obtain a non-integer rational point on the circle \(x^2+y^2=2\,\).
The point with coordinates \((p+\sqrt 2 \, q\,,\, r+\sqrt 2 \, s)\), where \(p\), \(q\), \(r\) and \(s\) are rational numbers, is called: an integer \(2\)-rational point if all of \(p\), \(q\), \(r\) and \(s\) are integers; a non-integer \(2\)-rational point if none of \(p\), \(q\), \(r\) and \(s\) is an integer.
\(\bf (a)\) Write down an integer \(2\)-rational point, and obtain a non-integer \(2\)-rational point, on the circle \(x^2+y^2=3\,\).
[\bf(b)] Obtain a non-integer \(2\)-rational point on the circle \(x^2+y^2=11\,\).
[\bf(c)]Obtain a non-integer \(2\)-rational point on the hyperbola \(x^2-y^2 =7 \).
Solution:
\(\bf (a)\) \((1, \sqrt2)\) is an integer \(2\)-rational point. \((\frac35 + \frac45\sqrt2, \frac45 - \frac{3}{5}\sqrt2)\) is a non-integer \(2\)-rational point.
[\bf(b)] First notice that \((\sqrt2)^2 +3^2 = 11\) so then consider \((1 + \tfrac32\sqrt2, 1-\tfrac32\sqrt2)\) will work as \(\pi/4\) degree rotation.
[\bf(c)] First notice \(3^2-(\sqrt2)^2 = 2\). Notice that \((k\sec \theta + \sqrt{m} \tan \theta)^2 - (k\tan \theta + \sqrt{m} \sec \theta)^2 = k^2-m\). Taking \(k= 3\) we have \((3 \cdot \frac{13}{5} + \frac{12}{5}\sqrt{2}, 3\cdot\frac{12}5+\frac{13}{5}\sqrt2)\)
Note: we can also find the additional point in the last part by considering lines through \((3, \sqrt2)\), for example \(y = -\frac32x + \sqrt2 + \frac92\) would give the same point.