# geometrical interpretations of the solutuon of equations- HELP PLEASE

• Oct 23rd 2007, 10:12 AM
lra11
geometrical interpretations of the solutuon of equations- HELP PLEASE
Hey this is probably really simple but I can't get my head around it.

1) Find the possible values of a for which y = ax+1 is a tangent to y = 3-4x+4

2) Find the possible values of k for which y = x-8 is a tangent to y= 4x²+ kx + 1

Would be really grateful if someone could just get me started and explain what to do. Thanks :)
• Oct 23rd 2007, 10:37 AM
Jhevon
Quote:

Originally Posted by lra11
Hey this is probably really simple but I can't get my head around it.

1) Find the possible values of a for which y = ax+1 is a tangent to y = 3-4x+4

I'll get you started with the first. the second should be similar.

what do we know about the line: $\displaystyle y = ax + 1$?

it has slope $\displaystyle a$ and y-intercept 1

how do we find the slope? by the derivative, so let's find the derivative. i will call the slope m.

$\displaystyle m = 6x - 4$ ..................(1)

ok, now we know that the given line passes through (0,1) and that all points on the curve are given by $\displaystyle \left( x, 3x^2 - 4x + 4 \right)$

thus, the tangent line must pass through the points $\displaystyle ( \underbrace{0}_{x_1} , \underbrace{1}_{y_1})$ and $\displaystyle ( \underbrace{x}_{x_2} , \underbrace{3x^2 - 4x + 4}_{y_2} )$

so another formula from the slope is (from precalc):

$\displaystyle m = \frac {y_2 - y_1}{x_2 - x_1} = \frac {3x^2 - 4x + 3}{x}$ ....................(2)

but we are talking about the same lines here, the slopes must be equal. thus we equate (1) and (2) to obtain:

$\displaystyle 6x - 4 = \frac {3x^2 - 4x + 3}x$

solving this, we find that $\displaystyle x = \pm 1$

thus, the points any tangent line would go through are:

$\displaystyle (0,1)$ and $\displaystyle (1,3)$ ....tangent line 1

or

$\displaystyle (0,1)$ and $\displaystyle (-1,11)$ ....tangent line 2

find the two lines that pass through those points and you will find the two values for $\displaystyle a$
• Oct 23rd 2007, 01:55 PM
ThePerfectHacker
Quote:

Originally Posted by lra11
Hey this is probably really simple but I can't get my head around it.

1) Find the possible values of a for which y = ax+1 is a tangent to y = 3-4x+4

For a parabola "being tangent" is the same as saying "only one intersection point".

Thus,
$\displaystyle ax+1 = 3x^2 - 4x+4$ has only a single solution.
Thus,
$\displaystyle 3x^2 - (4+a)x+3 = 0$
To have exactly one solution we require that,
$\displaystyle (4+a)^2 - 4(3)(3) = 0$
• Oct 23rd 2007, 01:58 PM
Jhevon
Quote:

Originally Posted by ThePerfectHacker
For a parabola "being tangent" is the same as saying "only one intersection point".

Thus,
$\displaystyle ax+1 = 3x^2 - 4x+4$ has only a single solution.
Thus,
$\displaystyle 3x^2 - (4+a)x+3 = 0$
To have exactly one solution we require that,
$\displaystyle (4+a)^2 - 4(3)(3) = 0$

well, that was easy. i need a Staples easy button
• Oct 24th 2007, 05:02 AM
lra11
thanks
thanks for the help both of you :)