Transform the equation xy-6x+2y+3=0 by translating the…

Transform the equation $xy-6x+2y+3=0$ by translating the origin to $O_1(-2,6)$. What curve is it?

Explanation of Method

Translating the origin to O₁(−2, 6) reveals an equilateral hyperbola.

Step 1: Substitution

Put $x=x'-2,\ y=y'+6$: $(x'-2)(y'+6)-6(x'-2)+2(y'+6)+3$.

\[ =x'y'+6x'-2y'-12-6x'+12+2y'+12+3=x'y'+15. \]

\[x'y'=-15\]

Step 2: Interpretation

It is an equilateral hyperbola whose asymptotes are the new axes, i.e. the lines $x=-2$ and $y=6$. Since $x'y'\lt 0$, the branches lie in the 2nd and 4th quadrants of the new system. Solving the original equation for $y$ confirms this:

\[ y=\frac{6x-3}{x+2}=6-\frac{15}{x+2}. \]

Check: the points $(1,1),(3,3),(-5,11),(-7,9)$ satisfy $xy-6x+2y+3=0$. In the new system they are $(3,-5),(5,-3),(-3,5),(-5,3)$, and each product is $-15$.

$x'$ $y'$ $(1,1)$ $(3,3)$ $(-5,11)$ $(-7,9)$ $O_1$

After moving the origin to $O_1(-2,6)$ the curve is $x'y'=-15$: an equilateral hyperbola whose asymptotes are the new axes. The marked points satisfy the original equation.

Result

\[\boxed{\,x'y'=-15\,}\]

Remark

Rotating by a further $45^\circ$ gives the canonical form $\dfrac{X^2}{30}-\dfrac{Y^2}{30}=-1$, so $a=b=\sqrt{30}$.