The curves y + ax^2 +b and y = 2x^2 +cx have a common tangent at the point (-1,0).

Find a, b and c.

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- Jun 13th 2006, 04:57 PMjayson81[SOLVED] Anyone know ?
The curves y + ax^2 +b and y = 2x^2 +cx have a common tangent at the point (-1,0).

Find a, b and c. - Jun 13th 2006, 06:03 PMThePerfectHackerQuote:

Originally Posted by**jayson81**

Substitute that into the second function,

$\displaystyle 0=2(-1)^2+c(-1)$

Thus,

$\displaystyle 0=2-c$ solve, $\displaystyle c=2$

Thus, the second function has equation,

$\displaystyle y=2x^2+2x$.

Since we say that these two function have a common tangent means the equation of the tangent through point (-1,0) is the same. Remember from analytic geomtery that two lines having same slope at same point are the same. Same thing here the tangent lines share a point thus they need to have the same slope.

The slope at $\displaystyle x=-1$ for the first function is,

$\displaystyle y'=2ax$ at $\displaystyle x=-1$ is $\displaystyle -2a$.

The slope for the second is,

$\displaystyle y'=4x+2$ at $\displaystyle x=-1$ is $\displaystyle -2$

Thus, $\displaystyle -2=-2a$ thus, $\displaystyle a=1$

All we need is to find "b" in,

$\displaystyle y=x^2+b$ since (-1,0) is on the curve we have,

$\displaystyle 0=(-1)^2+b$ thus, $\displaystyle b=-1$

Thus,

$\displaystyle (a,b,c)=(1,-1,2)$

Which is not true.

Thus, either you erred in the formulation of your question or there is no such numbers. - Jun 14th 2006, 05:08 AMSoroban
Hello, jayson81!

I got the same result as ThePerfectHacker, but I find that the solution works . . .

Quote:

The curves $\displaystyle y = ax^2 +b$ and $\displaystyle y = 2x^2 +cx$ have a common tangent at the point $\displaystyle (-1,0)$.

Find $\displaystyle a$, $\displaystyle b$ and $\displaystyle c$.

. . $\displaystyle a(-1)^2 + b \,=\,0\quad\Rightarrow\quad a + b \,=\,0$

. . $\displaystyle 2(-1)^2 + c(-1)\,=\,0\quad\Rightarrow\quad c = 2$

Since the slopes are equal at $\displaystyle (-1,0)$, we have:

. . $\displaystyle 2ax\,=\,4x + c\quad\Rightarrow\quad -2a\,=\,-4 + c$

Since $\displaystyle c = 2$, we get: $\displaystyle a = 1$ . . . and hence: $\displaystyle b = -1$

Therefore: .$\displaystyle (a,b,c)\,=\,(1,-1,2)$

The parabolas are: .$\displaystyle y\,=\,x^2 - 1$ and $\displaystyle y\,=\,2x^2 + 2x$

They share the common point $\displaystyle (-1,0)$ where they both have slope $\displaystyle -2.$