tangent line to 2y^3 + (6x^2)y - 12x^2 + 6y = 1

paulxzt

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Hi, I was wondering if anyone could help.

Consider the curve given by: 2y^3 + (6x^2)y - 12x^2 + 6y = 1

I solved the derivative which came out to be (4x-2xy)/(x^2 + y^2 + 1

1. Write an equation of each horizontal tangent line to the curve.
2. The line through the origin with slope -1 is tangent ot the curve at point P. Find the x- and y- coordinates of point P.

for 1 ) do I find when the derivative is 0? so I set the top = 0 to get y = 2.

for 2) im not even sure how to start this.. so I have m = -1 and point (0,0) .. now what
 
paulxzt said:
for 1 ) do I find when the derivative is 0? so I set the top = 0 to get y = 2.
How about x = 0, while you are at it?

for 2) im not even sure how to start this.. so I have m = -1 and point (0,0) .. now what
Where is the slope -1 and where does the line intersect the curve? If x = -y, that certainly simplifies the expression of the derivative.
 
Hello, paulxzt!

I can help with the Calculus, but the Algebra is baffling . . .


Consider the curve: 2y3+6x2y12x2+6y  =  1\displaystyle \,2y^3\,+\,6x^2y\,-\,12x^2\,+\,6y\;=\;1

1. Write an equation of each horizontal tangent line to the curve.

2. The line through the origin with slope -1 is tangent ot the curve at point P.
Find the x- and y- coordinates of point P.

I found the derivative: y=2x(2y)x2+y2+1  \displaystyle \:y' \:=\:\frac{2x(2\,-\,y)}{x^2\,+\,y^2\,+\,1}\; . . . Good!

(1) The derivative equals zero when its numerator is zero.

So we have: 2x(2y)=0\displaystyle \:2x(2\,-\,y) \:=\:0

And we have two cases: (a)y=2(b)x=0\displaystyle \,\begin{array}{cc}(a)\:y\,=\,2 \\ (b)\:x\,=\,0\end{array}

(a) Suppose y=2.\displaystyle y\,=\,2.

Substitute into the original equation: 223+6x2212x2+62  =  1\displaystyle \,2\cdot2^3\,+\,6x^2\cdot2\,-\,12x^2\,+\,6\cdot2 \;=\;1

And we have: 16+12x212x2+12=1        28=1  ??\displaystyle \,16\,+\,12x^2\,-\,12x^2\,+\,12 \:=\:1\;\;\Rightarrow\;\;28\:=\:1\;??

Hence: y2\displaystyle \,y\,\neq\,2


(b) Suppose x=0.\displaystyle x\,=\,0.

Substitute into the original equation: 2y3+602y1202+6y=1\displaystyle \,2y^3\,+\,6\cdot0^2\cdot y\,-\,12\cdot0^2\,+\,6y\:=\:1

And we have the cubic equation: 2y3+6y1  =  0\displaystyle \,2y^3\,+\,6y\,-\,1\;=\;0

But I see no simple way to solve for y.\displaystyle y.

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(2) I had better luck with this part . . .

The line through the origin with slope -1 has the equation: y=x\displaystyle \,y\:=\:-x

Substitute into the original equation: 2(x)3+6x2(x)12x2+6(x)  =  1\displaystyle \,2(-x)^3\,+\,6x^2(-x)\,-\,12x^2\,+\,6(-x)\;=\;1

And we get: 2x36x312x26x=1        8x3+12x2+6x+1=0\displaystyle \,-2x^3\,-\,6x^3\,-\,12x^2\,-\,6x\:=\:1\;\;\Rightarrow\;\;8x^3\,+\,12x^2\,+\,6x\,+\,1\:=\:0

I found that x=12\displaystyle x\,=\,-\frac{1}{2} is a root.
Using long division, I factored and got: (2x+1)3=0\displaystyle \,(2x\,+\,1)^3\:=\:0

So, x=12\displaystyle x\,=\,-\frac{1}{2} is the only root of the equation.
. . Hence, point P\displaystyle P has \(\displaystyle \fbox{x\,=\,-\frac{1}{2}}\)

Substitute into the original equation: 2y3+6(12)2y12(12)2+6y  =  1\displaystyle \,2y^3\,+\,6\left(-\frac{1}{2}\right)^2y\,-\,12\left(-\frac{1}{2}\right)^2\,+\,6y\;=\;1
. . and we have: 4y3+15y8=0\displaystyle \,4y^3\,+\,15y\,-\,8\:=\:0

I found that y=12\displaystyle y\,=\,\frac{1}{2} is a root.
. . So we have: (2y1)(2y2+y+8)=0\displaystyle \,(2y\,-\,1)(2y^2\,+\,y\,+\,8)\:=\:0

So y=12\displaystyle y\,=\,\frac{1}{2} is the only root of the equation.
. . Hence, point P\displaystyle P has \(\displaystyle \fbox{y\,=\,\frac{1}{2}}\)

Therefore: \(\displaystyle \,\fbox{P\left(-\frac{1}{2},\:\frac{1}{2}\right)}\)

 
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