Homographic function rotated by 45 degrees

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Consider the function $y=\dfrac{\sqrt2\,x+4}{x-\sqrt2}$:

• (a) write the function rotated by $45^\circ$ clockwise about the origin and find the semi-axes of the rotated conic (are they also those of the original conic?);
• (b) find vertices and foci of the rotated conic and deduce their positions for the original conic.

Solution

The function is a homographic hyperbola: $(x-\sqrt2)(y-\sqrt2)=6$, centre $C(\sqrt2;\sqrt2)$, $k=6>0$.

(a) It is rectangular: rotating by $45^\circ$ gives the canonical form $X^2-Y^2=2k=12$, i.e. $\dfrac{X^2}{12}-\dfrac{Y^2}{12}=1$. The semi-axes are $a=b=\sqrt{12}=2\sqrt3\approx3{,}46$. These are also the original conic’s semi-axes (rotation preserves size).

(b) In the rotated conic (centred at the origin): vertices $(\pm2\sqrt3;0)$, $c=2\sqrt6$, foci $(\pm2\sqrt6;0)$. For the original one, transferred around $C(\sqrt2;\sqrt2)$ along the bisector: vertices $(\sqrt2\pm\sqrt6;\sqrt2\pm\sqrt6)$, foci $(\sqrt2\pm2\sqrt3;\sqrt2\pm2\sqrt3)$.

$\boxed{a=b=2\sqrt3,\qquad k=6}$