Solving Tangent Line Problem: 2y^3 + (6x^2)y - 12x^2 + 6y = 1

[akxt]

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) I am not even sure how to start this.. so I have m = -1 and point (0,0) .. now what

 

[angel_of_pizza2000]

y = -x then is the line that is tangent to the original curve 2y^3 + (6x^2)y - 12x^2 + 6y = 1. -1 also equals that tangent, the derivative you obtained. The two equations - the derivative and the original curve - should have a common point, which is the point P you are looking for.

 

[HallsofIvy]

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.

Yes.

for 2) I am not even sure how to start this.. so I have m = -1 and point (0,0) .. now what

The line with m= -1, through (0,0) is, of course, y= -x. You now have two equations for the (x,y) coords of the point where that line is tangent to the curve: y'= (4x-2xy)/(x^2 + y^2 + 1)= -1 and the equation of the curve, 2y^3 + (6x^2)y - 12x^2 + 6y = 1. Knowing that y= -x should make that very easy. (If such a point exists.)

 

[surfsup887]

can someone do out part 2 of this problem?

 

[Defennder]

PF policy explicitly forbids doing homework for others. Just post any problems you encounter and we'll help.

 

[surfsup887]

i don't understand how to use both the equation and derivative to find the coordinate, can you explain how to relate the two?

 

[icystrike]

i don't understand how to use both the equation and derivative to find the coordinate, can you explain how to relate the two?

have you learned the topic differentiation?

 

[HallsofIvy]

i don't understand how to use both the equation and derivative to find the coordinate, can you explain how to relate the two?

You have two equations to solve for x and y. Solve the equations simultaneously.

 

[Darcey]

Do you mean set up the two equations equal to each other, then plug in -x for all y's then solve for x then plug that back into the equation to get the y?