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

• akxt
In summary, the conversation discusses finding horizontal tangent lines to a given curve and the coordinates of the point where a specific line is tangent to the curve. The process involves setting the derivative equal to 0 to find the y-coordinate, and then using that value to solve for the x-coordinate using the equation of the curve. The two equations are then solved simultaneously to find the coordinates of the point where the line is tangent to the curve.
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

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.

akxt said:
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.)

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can someone do out part 2 of this problem?

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

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

surfsup887 said:
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?

surfsup887 said:
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.

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?

## 1. What is a tangent line problem?

A tangent line problem involves finding the equation of a line that touches a given curve at a specific point. In this case, we are trying to find the equation of a line that touches the curve described by the equation 2y^3 + (6x^2)y - 12x^2 + 6y = 1 at a specific point.

## 2. How do you solve a tangent line problem?

To solve a tangent line problem, we need to find the slope of the curve at the given point and use it in the point-slope form of a line to find the equation of the tangent line. The slope of the curve at a point can be found using the derivative of the curve.

## 3. What is the point-slope form of a line?

The point-slope form of a line is y - y1 = m(x - x1), where m is the slope of the line and (x1, y1) is a point on the line. This form is useful when finding the equation of a line passing through a given point with a known slope.

## 4. How do you find the derivative of a curve?

The derivative of a curve can be found by using the power rule, product rule, quotient rule, or chain rule, depending on the form of the curve. In this problem, we will use the product rule to find the derivative of the given curve.

## 5. What are the steps to solving the given tangent line problem?

The steps to solving this tangent line problem are:
1. Find the derivative of the curve using the product rule.
2. Plug in the coordinates of the given point into the derivative to find the slope at that point.
3. Use the point-slope form of a line to find the equation of the tangent line, using the slope found in step 2 and the given point.
4. Simplify the equation of the tangent line, if necessary.
5. The final equation of the tangent line will be in the form y = mx + b, where m is the slope and b is the y-intercept.

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