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.

To find the horizontal tangent lines to the curve and the coordinates of point P, we need to set the derivative of y with respect to x (dy/dx) equal to zero and then solve for the values of x and y.

a) To find the horizontal tangent lines, we need to set dy/dx = 0.

Here's the equation again for reference: dy/dx = (4x - 2xy) / (x^2 + y^2 + 1)

Setting dy/dx equal to zero:

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

Now, we can multiply both sides of the equation by (x^2 + y^2 + 1) to eliminate the denominator:

0 = 4x - 2xy

Next, we can rearrange the equation to solve for y in terms of x:

2xy = 4x
y = 2

So, the equation of each horizontal tangent line is y = 2, as the value of y is constant regardless of the value of x.

b) Now let's find the coordinates of point P on the curve where the line through the origin with a slope of 0.1 is tangent to the curve.

The slope of the line through the origin is given as 0.1. We can write the slope-intercept form equation for the line as y = mx, where m is the slope.

So, the equation for the tangent line is y = 0.1x.

Now, we can substitute this equation into the original curve equation to find the coordinates of point P.

Substituting y = 0.1x into the original curve equation:

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

Simplifying the equation:

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

Combining like terms:

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

To solve this equation, you can use numerical methods like graphing calculators or computer software to find the approximate values of x and y.