common Tangents line to two curves asked by imad zak in many math groups

Exercise:

Consider a parabolas $\left( c \right):\,\,\,f\left( x \right)={{x}^{2}}\,\,\,\,\And \,\,\,\,\,\,\,\left( c' \right)\,\,:\,\,g\left( x \right)={{x}^{2}}-2x+2$

Determine the tangent equation to both curves .

Solution: we know that the tangent equation has the form $\left( d \right):\,\,\,\,y=ax+b$

So $\left( c \right)\cap \left( d \right)=\{I\}$ hence ${{x}^{2}}=ax+b\Leftrightarrow {{x}^{2}}-ax-b=0$

as $\left( d \right)$ tangent $\left( c \right)$ so we have double roots i.e $\Delta ={{a}^{2}}+4b=0$         (1)

Now $\left( d \right)\cap \left( c' \right)=\left\{ K \right\}$ hence $ax+b={{x}^{2}}-2x+2\Leftrightarrow {{x}^{2}}-\left( 2+a \right)x+2-b=0$

as $\left( d \right)$ tangent to $\left( c' \right)$ so we have double roots i.e $\Delta ={{\left( 2+a \right)}^{2}}-4\left( 2-b \right)=0$    (2)

So ${{a}^{2}}-{{\left( 2+a \right)}^{2}}=-8\Leftrightarrow -2\left( 2a+2 \right)=-4\left( a+1 \right)=-8\Leftrightarrow a=1$ thus $b=-\frac{1}{4}$

Or: Let $A\left( a,f\left( a \right) \right)$ be a on $\left( c \right)$ and $B\left( a',g\left( a' \right) \right)$

so ${{\left. \frac{df}{dx} \right|}_{x=a}}={{\left. 2x \right|}_{x=a}}=2a$\[{{\left. \frac{dg}{dx} \right|}_{x=a'}}={{\left( 2x-2 \right)}_{x=a'}}=2\left( a'-1 \right)\]

We are interesting in the parallel lines

So $2a=2\left( a'-1 \right)\Leftrightarrow a=a'-1\Leftrightarrow a'=a+1$

So the tangent lines at $A$ is :

$\left( d \right):\,\,y-{{y}_{A}}=f'\left( {{x}_{A}} \right)\left( x-{{x}_{A}} \right)\Leftrightarrow y-{{a}^{2}}=2a\left( x-a \right)\Leftrightarrow y=2ax-{{a}^{2}}$

Also the tangent line at $B$ is :

$\left( d' \right):\,\,y-{{y}_{B}}=g'\left( {{x}_{B}} \right)\left( x-{{x}_{B}} \right)\Leftrightarrow y=2\left( a'-1 \right)\left( x-a' \right)+a{{'}^{2}}-2a'+2$

$y=2a\left( x-a-1 \right)+{{\left( a+1 \right)}^{2}}-2\left( a'-1 \right)=2ax-2{{a}^{2}}-2a+{{a}^{2}}+2a+1-2a$

$\Leftrightarrow y=2ax-{{a}^{2}}-2a+1$ so ${{y}_{d}}={{y}_{d'}}\Leftrightarrow 2ax-{{a}^{2}}=2ax-{{a}^{2}}-2a+1\Leftrightarrow a=\frac{1}{2}\,\,\And \,\,a'=\frac{3}{2}$

So $\left( d \right):\,\,\,y=x-\frac{1}{4}$



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Idea of First method Credit to Imad Zak