Tangent lines

[Camillo]

Find equations of the lines that are tangent to both curves simultaneously :

$y = x^2+1 $
$y = -x^2 $.

[Cheguevilla]

In fact, I have sent you a PM with this "correction" days ago, Camillo.

I sent you a PM...

The period in which the event happens (days ago) is concluded.
The usage of present perfect is wrong.

[Camillo]

"Indeed" is stronger than "in fact ".
"Indeed " means in truth, really .
Example : are you sure of something ? Yes indeed, I am !

[fireball1]

On my discrete mathematics photocopies it's easy to find "in fact" instead of "indeed"!!!

[Camillo]

"fireball":In fact, I have sent you a PM with this "correction" days ago, Camillo.

Yes indeed, Fireball, but I forgot to make the correction

[Luca.Lussardi]

It is better to use indeed, instead of in fact.

[fireball1]

In fact, I have sent you a PM with this "correction" days ago, Camillo.

[Camillo]

"Marco83":Since we are in the mood for corrections:

Lines tangent ---> tangent lines

Correct

[Marco831]

Since we are in the mood for corrections:

Lines tangent ---> tangent lines

[_nicola de rosa]

"cheguevilla":

Making the same passages

Performing same steps...

curve of equation

curve with equation...

So the tangent lines are

Thus the tangent...

obtain

get...

Dear cheguevilla, many thanks for the corrections.
I hope not to have committed errors, now. Conversely, excuse me!

[Cheguevilla]

Making the same passages

Performing same steps...

curve of equation

curve with equation...

So the tangent lines are

Thus the tangent...

obtain

get...

[Camillo]

Correct .

[_nicola de rosa]

"Camillo":Find equations of the lines that are tangent to both curves simultaneously :

$y = x^2+1 $
$y = -x^2 $.

The equation of a tangent line is $y=mx+q$. We impose the condition of tangency with both the curves:
${(y=-x^2),(y=mx+q):}$ $->$ $x^2+mx+q=0$ and imposing $Delta=0$ we find $4q=m^2$
Making the same passages with the other curve of equazion $y = x^2+1 $ we have
${(y=x^2+1),(y=mx+q):}$ $->$ $x^2-mx+(1-q)=0$ and imposing $Delta=0$ we find $m^2-4+4q=0$
But $4q=m^2$, therefore we obtain $2m^2=4$, that is $m=+-sqrt(2)$ and $q=1/2$.
So the tangent lines are
$y=sqrt(2)x+1/2$ and $y=-sqrt(2)x+1/2$