Consider the curve x^{2}y+y^{2}x=6 A) Find \frac {dy}{dx} in terms of x and y B) Write the equation for the tangent line where x=2 and y=1. C) Find the coordinates of the point (x,y) where the curve has a horizontal tangent line.

A) To find dy/dx in terms of x and y, we can use implicit differentiation.

First, let's differentiate both sides of the equation with respect to x:

d/dx (x^2y + y^2x) = d/dx (6)

Using the product rule, we get:

2xy + x^2(dy/dx) + 2yx + y^2 = 0

Simplifying this equation, we have:

2xy + 2yx + x^2(dy/dx) + y^2 = 0

Combining like terms, we get:

4xy + x^2(dy/dx) + y^2 = 0

Now, let's isolate dy/dx:

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

So, dy/dx in terms of x and y is (-4xy - y^2) / x^2.

B) To find the equation of the tangent line where x = 2 and y = 1, we can use the point-slope form of a line.

First, let's find the slope of the tangent line by substituting x = 2 and y = 1 into the expression we found for dy/dx:

dy/dx = (-4(2)(1) - (1)^2) / (2)^2 = (-8 - 1) / 4 = -9/4

So, the slope of the tangent line is -9/4.

Now, let's use the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point (2, 1) and m is the slope we found:

y - 1 = (-9/4)(x - 2)

Simplifying this equation, we get:

y - 1 = (-9/4)x + 9/2

Adding 1 to both sides, we have:

y = (-9/4)x + 9/2 + 1 = (-9/4)x + 11/2

So, the equation of the tangent line where x = 2 and y = 1 is y = (-9/4)x + 11/2.

C) To find the coordinates of the point (x, y) where the curve has a horizontal tangent line, we need to find the value(s) of x where dy/dx = 0.

From part A, we have:

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

For a horizontal tangent line, dy/dx = 0. So, we can set the numerator equal to 0:

-4xy - y^2 = 0

Factoring out y, we get:

y(-4x - y) = 0

This equation is satisfied when either y = 0 or -4x - y = 0.

If y = 0, then the curve intersects the x-axis. To find the x-coordinate(s), we can substitute y = 0 into the original equation:

x^2(0) + (0)^2x = 6 0 + 0 = 6

This equation is not satisfied, so there are no points on the curve where y = 0.

If -4x - y = 0, then we can solve for y in terms of x:

-4x - y = 0 y = -4x

Substituting this into the original equation, we get:

x^2(-4x) + (-4x)^2x = 6 -4x^3 - 16x^3 = 6 -20x^3 = 6 x^3 = -6/20 x^3 = -3/10

Taking the cube root of both sides, we have:

x = -∛(3/10)

So, the x-coordinate of the point where the curve has a horizontal tangent line is x = -∛(3/10).

To find the corresponding y-coordinate, we can substitute this value of x into the equation y = -4x:

y = -4(-∛(3/10)) y = 4∛(3/10)

Therefore, the coordinates of the point (x, y) where the curve has a horizontal tangent line are x = -∛(3/10) and y = 4∛(3/10).