Find Tangent Line to Curve (x+2y)^2+2x-y-3=0: Answer at Yahoo Answers

[MarkFL]

Here is the question:

Find the equation of the tangent line to the curve (x+2y)^2+2x-y-3+0 parallel to 4x+3y=2?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Vina,

I am assuming the curve is defined as follows:

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

Implicitly differentiating with respect to $x$, we obtain:

\(\displaystyle 2(x+2y)(1+2y')+2-y'=0\)

Distribute:

\(\displaystyle 2(x+2y)+4(x+2y)y'+2-y'=0\)

Move everything that does not have $y'$ as a factor to the right side:

\(\displaystyle 4(x+2y)y'-y'=-\left(2(x+2y)+2 \right)\)

Factor both sides:

\(\displaystyle \left(4(x+2y)-1 \right)y'=-2\left(x+2y+1 \right)\)

Divide through by \(\displaystyle 4(x+2y)-1\):

\(\displaystyle y'=-\frac{2\left(x+2y+1 \right)}{4(x+2y)-1}\)

Now, we want to equate this to the slope $m$ of the given line:

\(\displaystyle 4x+3y=2\)

which we see is:

\(\displaystyle m=-\frac{4}{3}\)

And so we have:

\(\displaystyle \frac{2\left(x+2y+1 \right)}{4(x+2y)-1}=\frac{4}{3}\)

Dividing through by 2 and cross-multiplying yields:

\(\displaystyle 3x+6y+3=8x+16y-2\)

Collect like terms:

\(\displaystyle 5=5x+10y\)

Divide through by 5 and arrange as:

\(\displaystyle x=1-2y\)

Now, substituting for $x$ into the original equation, we obtain:

\(\displaystyle (1-2y+2y)^2+2(1-2y)-y-3=0\)

Solving for $y$, we find:

\(\displaystyle 1+2-4y-y-3=0\)

\(\displaystyle y=0\implies x=1\)

Now, we have the point of tangency $(1,0)$ and the slope \(\displaystyle m=-\frac{4}{3}\), and so the point-slope formula gives is the equation of the tangent line:

\(\displaystyle y-0=-\frac{4}{3}(x-1)\)

In slope-intercept form, this is:

\(\displaystyle y=-\frac{4}{3}x+\frac{4}{3}\)

Here is a plot of the given curve, line and the tangent line:

View attachment 2109

Here is a link to the program I used to plot them:

(x+2y)^2+2x-y-3=0,y=(4/3)(1-x),y=(2-4x)/3, where x=0 to 2,y=-1 to 1 - Wolfram|Alpha