How Do You Find a Common Tangent Line to Two Parabolas?

[karush]

https://www.physicsforums.com/attachments/2227
The region $R$, is bounded by the graphs of
\(\displaystyle f(x)=x^2 -3\) and \(\displaystyle g(x)=(x-3)^2\),
and the line \(\displaystyle T\), as is shown in the figure above.
\(\displaystyle T\) is tangent to the graph of f at point \(\displaystyle (a,a^2-3) \)
and tangent to the graph of g at point \(\displaystyle (b,(b-3)^2)\)

a. Show that $a=b-3$
b. Show the numerical value of $a$ and $b$
c. Write the equation of $T$
d. Setup but do not calculate the integral for the region $R$.

this is a common problem, and I looked at several solutions via several forums but only got lost on this. I know that the slope of the $2$ parabola's is $2x$ and since it is a line the slope is the same. I did try using $y=mx+b$ but with $3$ variables $a,b$ and $x$ it was hard work with. If I could see how the equation of $T$ is derived then the other questions I should be able to answer.
the answer is not given but the line eq appears to be close to $y=x-3.25$

 

[MarkFL]

Re: tangent line to 2 parabola’s

What I would for part c) is let the tangent line be $y=mx+b$. Next, equate this line to the two parabolas in turn and require the resulting discriminants to both be zero, giving two equations in two unknowns from which $m$ and $b$ can be easily found. Once you have this, then the other parts of the question should be a piece of cake. :D

 

[karush]

Re: tangent line to 2 parabola’s

not sure if I follow ...
but presume $T$ is both $2x(x-a)+(a^2-3)$ and $2x(x-b)+(b-3)^2$

 

[MarkFL]

Re: tangent line to 2 parabola’s

What I did was:

i) \(\displaystyle x^2-3=mx+b\)

standard form...

\(\displaystyle x^2-mx-(b+3)=0\)

discriminant to zero...

\(\displaystyle m^2+4b+12=0\)

ii) \(\displaystyle (x-3)^2=mx+b\)

standard form...

\(\displaystyle x^2-(m+6)x+(9-b)=0\)

discriminant to zero...

\(\displaystyle m^2+12m+4b=0\)

Now, since both discriminants are zero, we may equate them and write:

\(\displaystyle m^2+4b+12=m^2+12m+4b\)

Collect like terms:

\(\displaystyle 12m=12\)

Divide through by 12:

\(\displaystyle m=1\)

And thus, we can find from either equation:

\(\displaystyle b=-\frac{13}{4}\)

Hence, the tangent line is:

\(\displaystyle y=x-\frac{13}{4}\)

 

[karush]

Re: tangent line to 2 parabola’s

OK, I see now, will try to get the rest of it. that was a big help
BTW I have an incredible view of Pearl Harbor here at LCC