original curve: 2y^3+6(x^2)y-12x^2+6y=1

dy/dx=(4x-2xy)/(x^2+y^2+1)

a) write an equation of each horizontal tangent line to the curve
b) the line through the origin with the slope .1 is tangent to the curve at P. Find x and y of point P.

if dy/dx is zero, that must mean 4x-2xy is zero, or y=2

When y=2, solve for x in the original equation.

when you plug it in the original eqn. it doesn't work

28=1 is the value of x and that doesn't work

To find the equation of each horizontal tangent line to the curve, we need to find the points on the curve where the derivative dy/dx equals zero.

First, let's find the derivative dy/dx of the original curve. The given derivative is:

dy/dx = (4x - 2xy) / (x^2 + y^2 + 1)

Now, set dy/dx equal to zero and solve for x and y:

0 = (4x - 2xy) / (x^2 + y^2 + 1)

To make it easier to solve, let's multiply both sides by (x^2 + y^2 + 1):

0 = 4x - 2xy

Now, we can isolate y:

2xy = 4x

y = 2

So, for each horizontal tangent line, the value of y will be 2.

Thus, the equation of each horizontal tangent line is y = 2.

Now, let's move on to part b.

We want to find the point P on the curve where the line through the origin with a slope of 0.1 is tangent to the curve.

The equation of the line through the origin with slope 0.1 can be written as:

y = 0.1x

Now, substitute this equation into the original curve:

2(0.1x)^3 + 6(x^2)(0.1x) - 12x^2 + 6(0.1x) = 1

Simplify the equation:

0.002x^3 + 0.06x^3 - 12x^2 + 0.6x - 1 = 0

Combine like terms:

0.062x^3 - 12x^2 + 0.6x - 1 = 0

Now, we can use numerical methods or a graphing calculator to solve this equation for x. Let's assume the solution is x = a.

Substitute this value back into the equation of the line:

y = 0.1(a)

So, the x-coordinate of point P is a, and the y-coordinate of point P is 0.1a.

To find the exact values of x and y, you will need to solve the equation 0.062x^3 - 12x^2 + 0.6x - 1 = 0 using appropriate numerical methods or a graphing calculator.