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Considering the function,

$$y=\frac{ax+b}{cx+d}\tag1$$ If $c = 0 \wedge d\neq 0$, the function represents a straight line of equation

$$y=\frac ad x+ \frac bd$$

If $c ≠ 0$ and $ad = bc$ the function represents a horizontal straight line. In fact, if

$$ad = bc \tag 2$$

we will have

$$ad/c = bc/c \iff ad/c = b$$

The coordinates of the point $P_0(-d/c,a/c)$ represent the asyntots of our hyperbola $(1)$. The importance of $(2)$ is due to the reason that if $ad-bc \neq 0$, using the traslation $\tau$, $$\tau: \begin{cases} X=x+\dfrac dc & \\ Y=y-\dfrac ac \end{cases} $$

I will obtain an equilater hyperbola. In fact

$$Y+\frac{a}{c}=\frac{a\Big(X-\frac{d}{c}\Big)+b}{c\Big(X-\frac{d}{c}\Big)+d}$$

$$Y=\frac{aX-\frac{ad}{c}+b}{cX-d+d}-\frac{a}{c}\Rightarrow Y=\frac{aX-\frac{ad}{c}+b}{cX}-\frac{a}{c}\Rightarrow Y=\frac{aX-\frac{ad}{c}+b-aX}{cX}$$

Hence:

$$Y=\frac{-\frac{ad}{c}+b}{cX}\Rightarrow XY=-\frac{ad}{c^2}+\frac{b}{c}\Rightarrow XY=k$$ with $$k=\frac{bc-ad}{c^2}$$

$$XY=k \tag 3$$

Starting from $(1)$ how can I create the condition quickly (step by step) $$\boxed{\color{orange}{ad-bc}} \quad ?$$ different from my proof?

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  • $\begingroup$ I'm not sure to have understood precisely your point. From here we have $$y=\frac{ax+b}{cx+d}=\frac{c(ax+b)}{c(cx+d)}=\frac{a(cx+\frac{bc}a)}{c(cx+d)}$$ which is constant when $\frac{bc}a=d$. $\endgroup$ – user Jul 31 at 23:01
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    $\begingroup$ @user Not really, I'd like to get the $ad-bc$ condition starting from the $(1)$. $\endgroup$ – Sebastiano Jul 31 at 23:06
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If $c\neq 0$, then $$y=\frac ac+\left(\frac{ax+b}{cx +d}-\frac ac\right)=\frac ac+\frac{bc-ad}{c(cx +d)}.$$

If $d\neq 0$, then $$y=\frac bd+\left(\frac{ax+b}{cx+d}-\frac bd\right)=\frac bd+\frac{(ad-bc)x}{d(cx+d)}.$$

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    $\begingroup$ Very very nice...I have appreciated very lot. Best regards....and thankkkkk youuuuuuuu. $\endgroup$ – Sebastiano Aug 4 at 9:24

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