Consider the curve y^2+xy+x^2=15. What is dy/dx?

Find the two points on the curve where y equals 0 and show that the tangent lines there are parallel.

Find the coordinates of all points on the curve where there is a horizontal tangent line, with work.

To find dy/dx, we differentiate the equation of the curve implicitly with respect to x.

Let's start by differentiating both sides of the equation y^2 + xy + x^2 = 15 with respect to x.

On the left side, we have to apply the chain rule. Differentiating y^2 with respect to x gives 2yy'. Differentiating xy gives x(dy/dx) + y, and differentiating x^2 gives 2x.

So the left side becomes 2yy' + x(dy/dx) + y + 2x.

On the right side, the derivative of a constant is zero, so it becomes 0.

Now we can simplify the equation:

2yy' + x(dy/dx) + y + 2x = 0

To find dy/dx, we want to isolate it. Rearrange the equation to solve for dy/dx:

x(dy/dx) + 2yy' = -y - 2x

Now, factor out dy/dx:

dy/dx(x + 2y) = -y - 2x

Finally, divide both sides by (x + 2y):

dy/dx = (-y - 2x) / (x + 2y)

So, the derivative of y with respect to x, dy/dx, is given by (-y - 2x) / (x + 2y).

Now let's find the two points on the curve where y equals 0. We substitute y = 0 into the equation y^2 + xy + x^2 = 15:

0 + 0 + x^2 = 15

Simplifying, we have x^2 = 15. Taking the square root of both sides, we get x = ±√15.

Therefore, the two points on the curve where y equals 0 are (-√15, 0) and (√15, 0).

To show that the tangent lines at these points are parallel, we can compare their slopes. The slope of a tangent line is given by dy/dx. Let's calculate dy/dx at each of the points.

For the point (-√15, 0):

dy/dx = (-y - 2x) / (x + 2y)
= (-(0) - 2(-√15)) / (-√15 + 0)
= (2√15) / (-√15)
= -2

For the point (√15, 0):

dy/dx = (-y - 2x) / (x + 2y)
= (-(0) - 2(√15)) / (√15 + 0)
= (-2√15) / (√15)
= -2

Both calculations give us a slope of -2. Since the slopes are equal, this implies that the tangent lines at (-√15, 0) and (√15, 0) are parallel.

Finally, let's find the coordinates of all points on the curve where there is a horizontal tangent line.

To have a horizontal tangent line, dy/dx must equal zero. So we set (-y - 2x) / (x + 2y) = 0.

Simplifying this expression, we get -y - 2x = 0.

Rearranging the equation, we have y = -2x.

Substituting this expression for y into the equation y^2 + xy + x^2 = 15, we get (-2x)^2 + x(-2x) + x^2 = 15.

Simplifying, we have 4x^2 - 2x^2 + x^2 = 15.

Combining like terms, we get 3x^2 = 15.

Dividing both sides by 3, we find x^2 = 5.

Taking the square root of both sides, we have x = ±√5.

So the coordinates of all points on the curve where there is a horizontal tangent line are (√5, -2√5) and (-√5, 2√5).

Therefore, the points on the curve where there is a horizontal tangent line are (√5, -2√5) and (-√5, 2√5).

To find dy/dx for the given curve, we need to differentiate the equation with respect to x implicitly.

Differentiating y^2 + xy + x^2 = 15 with respect to x, we get:
2y(dy/dx) + y + x(dy/dx) + 2x = 0

Rearranging the terms, we can solve for dy/dx:
(dy/dx)(2y + x) = - (2x + y)
dy/dx = -(2x + y) / (2y + x)

Now, let's find the points on the curve where y equals 0:
Substitute y = 0 into the equation y^2 + xy + x^2 = 15:
(0)^2 + x(0) + x^2 = 15
x^2 = 15
x = ±√15

Therefore, the two points on the curve where y equals 0 are (-√15, 0) and (√15, 0).

To show that the tangent lines at these points are parallel, we can calculate the slopes of the tangent lines at each point. If the slopes are equal, then the lines are parallel.

At point (-√15, 0):
Substituting x = -√15 and y = 0 into dy/dx:
dy/dx = -(2(-√15) + 0) / (2(0) + (-√15))
dy/dx = √15 / -√15
dy/dx = -1

At point (√15, 0):
Substituting x = √15 and y = 0 into dy/dx:
dy/dx = -(2(√15) + 0) / (2(0) + √15)
dy/dx = -√15 / √15
dy/dx = -1

Both slopes are -1, which means the tangent lines at (-√15, 0) and (√15, 0) are parallel.

Now, let's find the coordinates of all points on the curve where there is a horizontal tangent line. For a horizontal tangent line, dy/dx must be equal to 0.

Setting dy/dx = 0 in the equation -(2x + y) / (2y + x) = 0, we get:
2x + y = 0

Substituting y = -2x into the equation y^2 + xy + x^2 = 15, we have:
(-2x)^2 + x(-2x) + x^2 = 15
4x^2 - 2x^2 + x^2 = 15
3x^2 = 15
x^2 = 5
x = ±√5

Therefore, the coordinates of the points on the curve where there is a horizontal tangent line are (√5, -2√5) and (-√5, 2√5).