Show that the tangents to the curve y = (x-4)/(x-2)
Show that the tangents to the curve $y = \dfrac{x-4}{x-2}$ at its intersections with the coordinate axes are parallel.
Explanation of Method
Two intercepts equidistant from the pole, hence equal slopes.
Step 1: The two intercepts
\[ y = 0 \;\Longrightarrow\; x = 4, \qquad A = \bigl(4,0\bigr); \qquad x = 0 \;\Longrightarrow\; y = \frac{-4}{-2} = 2, \qquad B = \bigl(0,2\bigr). \]
Step 2: The derivative
\[ y' = \frac{\bigl(x-2\bigr)-\bigl(x-4\bigr)}{\bigl(x-2\bigr)^{2}} = \frac{2}{\bigl(x-2\bigr)^{2}}. \tag{1} \]
Step 3: The two slopes
The points $x = 4$ and $x = 0$ are symmetric about the pole $x = 2$, so $\bigl(x-2\bigr)^{2}$ takes the same value $4$ at both: \[k_{A} = \frac{2}{4} = \frac12 = \frac{2}{4} = k_{B}.\]
Step 4: The tangents
\[ \text{at } A:\ y = \frac{x}{2}-2; \qquad \text{at } B:\ y = \frac{x}{2}+2. \]
Equal slopes and different intercepts: the lines are parallel and distinct.
Step 5: Verification
\[\text{At } A:\ \frac42-2 = 0 \;\;\checkmark; \qquad \text{at } B:\ 0+2 = 2 \;\;\checkmark\]
\[y' \gt 0 \text{ everywhere: the curve increases on each branch, so equal slopes are possible only in this symmetric way} \;\;\checkmark\]
\[\text{Centre of symmetry } \bigl(2,1\bigr):\ y = 1+\frac{-2}{x-2},\ \text{and } A,B \text{ are symmetric about it} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,k_{A} = k_{B} = \frac12 \;\Longrightarrow\; \text{the tangents } y = \frac{x}{2}-2 \text{ and } y = \frac{x}{2}+2 \text{ are parallel}\,}\]
Remark
Remark. Writing the function as $y = 1-\dfrac{2}{x-2}$ exhibits the centre of symmetry $\bigl(2,1\bigr)$; the two intercepts $A\bigl(4,0\bigr)$ and $B\bigl(0,2\bigr)$ are images of each other under the half-turn about that centre, and a half-turn carries tangents to parallel tangents. The computation merely confirms what the symmetry already guarantees.