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

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To find the tangent line to the curve defined by (x+2y)^2 + 2x - y - 3 = 0 that is parallel to the line 4x + 3y = 2, implicit differentiation is performed to derive the slope of the curve. The slope of the given line is determined to be -4/3. By setting the derivative equal to this slope, the equation simplifies to find the point of tangency, which is calculated to be (1, 0). The equation of the tangent line is then derived using the point-slope formula, resulting in y = -4/3x + 4/3. A plot of the curve, the line, and the tangent line is provided for visual reference.
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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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