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

[paulxzt]

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

 

[tkhunny]

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.

 

[soroban]

Hello, paulxzt!

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

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

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: \(\displaystyle \:y' \:=\:\frac{2x(2\,-\,y)}{x^2\,+\,y^2\,+\,1}\;\) . . . Good!

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

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

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

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

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

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

Hence: \(\displaystyle \,y\,\neq\,2\)


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

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

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

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

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

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

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

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

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

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

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

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

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

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