Find al points where tangent lines to: (x^2+y^2)^(3/2) = sqrt(x^2+y^2) + x is either horizontal or vertical.

To find the points where the tangent lines to the equation (x^2+y^2)^(3/2) = sqrt(x^2+y^2) + x are either horizontal or vertical, we need to find the values of x and y which satisfy this condition.

Let's start by differentiating both sides of the equation with respect to x to find the slope of the tangent line at any point (x, y):

Using the chain rule, we differentiate (x^2+y^2)^(3/2) with respect to x:

d/dx((x^2+y^2)^(3/2)) = d/dx(sqrt(x^2+y^2) + x)

To differentiate the left side, we use the chain rule:

(3/2)(x^2+y^2)^(1/2) * d/dx(x^2+y^2) = d/dx(sqrt(x^2+y^2) + x)

Simplifying the left side:

(3/2)(x^2+y^2)^(1/2) * (2x + 2y * dy/dx) = 1/sqrt(x^2+y^2) + 1

Now, let's solve for dy/dx to find the slope of the tangent line at any point (x, y):

(3/2)(x^2+y^2)^(1/2) * (2x + 2y * dy/dx) = 1/sqrt(x^2+y^2) + 1

Rearranging the equation:

dy/dx = (1/sqrt(x^2+y^2) + 1 - (3/2)(x^2+y^2)^(1/2) * 2x) / (2y * (3/2)(x^2+y^2)^(1/2))

To find the points where the tangent lines are either horizontal or vertical, we need the slope (dy/dx) to be either 0 (for horizontal tangent lines) or undefined (for vertical tangent lines).

First, let's find the points where the slope is 0 (horizontal tangent lines):

Setting dy/dx = 0:

0 = (1/sqrt(x^2+y^2) + 1 - (3/2)(x^2+y^2)^(1/2) * 2x) / (2y * (3/2)(x^2+y^2)^(1/2))

Multiplying both sides by 2y * (3/2)(x^2+y^2)^(1/2):

0 = 1/sqrt(x^2+y^2) + 1 - (3/2)(x^2+y^2)^(1/2) * 2x

Rearranging the equation:

1 = (3/2)(x^2+y^2)^(1/2) * 2x - 1/sqrt(x^2+y^2)

To simplify the equation further and find the points, we need to use algebraic techniques to isolate x and y. However, due to the complexity of the equations involved, it is not possible to find a simple solution analytically.

To find the points where the tangent lines are vertical, we need to find where the slope is undefined. This occurs when the denominator of dy/dx is equal to 0, which means 2y * (3/2)(x^2+y^2)^(1/2) = 0.

Setting 2y * (3/2)(x^2+y^2)^(1/2) = 0:

2y = 0

This means y = 0.

Therefore, the equation (x^2+y^2)^(3/2) = sqrt(x^2+y^2) + x has vertical tangent lines at points where y = 0.

In summary, the equation has vertical tangent lines when y = 0, but finding the points where the tangent lines are horizontal requires further algebraic manipulation and is not easily solvable analytically.