# Finding points where a tangent lines is parallel to another line

• Oct 15th 2009, 06:32 AM
hazecraze
Finding points where a tangent lines is parallel to another line
#8. find each point at which the tangent line to the curve $\displaystyle y1=2x +(4/x) +1$ is parallel to the line $\displaystyle y2+2x=6$.

How would I do this? Take y1' and get y1'(4) and set it equal to the slope of y2 and then plug in y1 for x and y values? The answer is (1, 7) and (-1, -5).
• Oct 15th 2009, 06:42 AM
stapel
Solve the second equation for "y=". Then read the slope value off the equation. (It'll be the value multiplied on the "x".)

Differentiate the first equation with respect to x. Set equal to the given slope value, and solve for the corresponding x-value(s).

Plug these x-values into the first equation, and solve for the corresponding y-value(s), and thus the point(s) in question.

(I get the same answer as you've listed.) (Wink)
• Oct 15th 2009, 08:04 AM
hazecraze
At first I confused my self by thinking it was $\displaystyle y1=2x +(4/x +1)$ and got (1,3), but I see that it was $\displaystyle y1=2x +(4/x) +1$. Also, after differentiating and setting equal to -2 and then simplifying, I got $\displaystyle -(x^3+2x^2+x+1)$. With synthetic division and a $\displaystyle (x-1)$ root, I got -(x-1)=0, which was the $\displaystyle x$ value of (1,7), but for the other part: $\displaystyle (x^2+x)$, I factored out an $\displaystyle x$ to get $\displaystyle x(x+1)$, so wouldn't there also be a point 0, in addition to the -1, thus giving (0,1) and then the (1,7) and (-1,-5)?
• Oct 15th 2009, 08:40 AM
ramiee2010
Quote:

Originally Posted by hazecraze
#8. find each point at which the tangent line to the curve $\displaystyle y1=2x +(4/x) +1$ is parallel to the line $\displaystyle y2+2x=6$.

How would I do this? Take y1' and get y1'(4) and set it equal to the slope of y2 and then plug in y1 for x and y values? The answer is (1, 7) and (-1, -5).

$\displaystyle slop\ of\ line \ y_{2}+2x=6 is m_{2}=-2\quad (\because y=mx+c)$
$\displaystyle slop\ of \ curve \ y_{1}= 2x +\frac{4}{x} +1 \ is$
$\displaystyle m_{1}=\frac{dy}{dx}=\quad2- \frac{4}{x^2}$
$\displaystyle since\ the\ tangent\ line\ y_{1}\ to\ the\ curve\ is\ parallel\ to\ the\ line\ y_{2}$
$\displaystyle m_{1}=m_{2} \quad \ so \ that\ \quad \quad2- \frac{4}{x^2} = -2$
$\displaystyle or \ x= \pm 1$
on substituting in equation of curve we get
$\displaystyle for \ x=1 \quad y=7 \ and\ for \ x=-1 \quad y=-5$
at (1, 7) and (-1, -5) ,the tangent line to the curve $\displaystyle \ y_{1}= 2x +\frac{4}{x} +1$ is parallel to the line $\displaystyle y_{2}+2x=6$http://img387.imageshack.us/img387/3531/graph.jpg
• Oct 15th 2009, 09:31 AM
hazecraze
Oh, well that was much more simplistic than what I did.