Q.Find the equation of the rectangular hyperbola which has for one of its asymptotes the line $x + 2y - 5 = 0$ and passes through the points $(6, 0)$ and $(-3, 0)$.

Use perpendicularity of a rectangular hyperbola's asymptotes to find the second asymptote, then fit the pencil $L_1L_2=k$ to the two given points.

1. Key property. A rectangular hyperbola is one whose asymptotes are mutually perpendicular. One asymptote is given as $L_1: x+2y-5=0$, i.e. $y=-\tfrac12x+\tfrac52$, which has slope $m_1=-\tfrac12$.

2. Find the second asymptote. Perpendicularity requires $m_1m_2=-1$, so $m_2 = 2$. Hence the second asymptote has the form

   $L_2: y = 2x+d \quad\text{i.e.}\quad 2x-y+c=0,$for some constant $c$ to be determined (here $c=-d$).

3. General equation through a pair of asymptotes. Any conic having $L_1=0$ and $L_2=0$ as its asymptotes can be written as

   $L_1\cdot L_2 = k \quad\text{i.e.}\quad (x+2y-5)(2x-y+c)=k,$for constants $c,k$ (this is the standard rectangular-hyperbola-from-asymptotes form, since $L_1L_2=0$ is exactly the degenerate pair of asymptotes and adding a constant shifts it off the asymptotes onto the actual hyperbola).

4. Use the point $(6,0)$.

   $(6+0-5)(12-0+c)=k \implies (1)(12+c)=k \implies 12+c=k. \quad (i)$

5. Use the point $(-3,0)$.

   $(-3+0-5)(-6-0+c)=k \implies (-8)(c-6)=k \implies 48-8c=k. \quad (ii)$

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