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.