SOLUTION: Find the equation of the tangent to the curve (x+2y)^2+2x-y-3=0 parallel the line 4x+3y=12.

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Question 1066886: Find the equation of the tangent to the curve (x+2y)^2+2x-y-3=0 parallel the line 4x+3y=12.
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

The red curve is the graph of the equation %28x%2B2y%29%5E2%2B2x-y-3=0
The black line is the graph of the equation 4x%2B3y=12 



Solve 4x%2B3y=12 for y:

       y=expr%28-4%2F3%29x%2B4

The slope of any line parallel to it has slope -4%2F3

and the equation of any line parallel to y=expr%28-4%2F3%29x%2B4

has equation y=expr%28-4%2F3%29x%2Bb for some value of b

So we solve the system by substitution: 

system%28%28x%2B2y%29%5E2%2B2x-y-3=0%2Cy=expr%28-4%2F3%29x%2Bb%29

You do that and solve for x.  I did it on paper, but I don't 
feel like typing all the steps in.  If you can't get it, tell 
me in the thank-you note form below and I'll help you do it.
It involves the quadratic formula.

Here is what I got for the x values of the two solutions:

x=expr%283%2F5%29%28b-1+%2B-+sqrt%284-3b%29%29

The line will be tangent to the curve if both solutions are
equal.  We see that they will be equal if the square roots are 
both 0.

So we set what's under the square root equal to 0:

4-3b=0

b=+4%2F3

So the equation of the line tangent to the curve parallel to

4x+3y=12

is   y=expr%28-4%2F3%29x%2B4%2F3

which in general form is

4x%2B3y=4

We draw it in green:



It turns out that the line is tangent at the point (1,0).

Edwin