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

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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.
 
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Hello Vina,

I am assuming the curve is defined as follows:

$$(x+2y)^2+2x-y-3=0$$

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

$$2(x+2y)(1+2y')+2-y'=0$$

Distribute:

$$2(x+2y)+4(x+2y)y'+2-y'=0$$

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

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

Factor both sides:

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

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

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

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

$$4x+3y=2$$

which we see is:

$$m=-\frac{4}{3}$$

And so we have:

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

Dividing through by 2 and cross-multiplying yields:

$$3x+6y+3=8x+16y-2$$

Collect like terms:

$$5=5x+10y$$

Divide through by 5 and arrange as:

$$x=1-2y$$

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

$$(1-2y+2y)^2+2(1-2y)-y-3=0$$

Solving for $y$, we find:

$$1+2-4y-y-3=0$$

$$y=0\implies x=1$$

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

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

In slope-intercept form, this is:

$$y=-\frac{4}{3}x+\frac{4}{3}$$

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
 

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