find the points on the curve x^2+xy+y^2=7 (a) where the tangent is parallel to the x-axis and (b) where the tangent is parallel to the y-axis. In the latter case, dy⁄dx is not defined, but dx⁄dy is. What value does dx⁄dy have at these points?

to find dy/dx,

2x + y + xy' + 2yy' = 0
y' = -(2x+y)/(x+2y)
y'=0 where 2x+y=0

x^2 + x(-2x) + (-2x)^2 = 7
x^2 - 2x^2 + 4x^2 = 7
3x^2 = 7
x = √(7/3)
y = -2√*7/3)

interchange the variables for dx/dy

naturally, when dy/dx=0, dx/dy is undefined, and vice-versa.

To see this all laid out, visit

wolframalpha.com and enter

plot x^2+xy+y^2=7, y^2 = 28/3, x^2 = 28/3

To find the points on the curve where the tangent is parallel to the x-axis, we need to solve for dy/dx = 0.

Given the equation of the curve: x^2 + xy + y^2 = 7

(a) Tangent parallel to the x-axis:
To find the points where the tangent is parallel to the x-axis, we need to find the values of x and y that satisfy dy/dx = 0.

Differentiating the equation implicitly with respect to x:
2x + (xy + y^2)dx/dx + x(dy/dx) + 2y(dy/dx) = 0

Simplifying the equation:
2x + (xy + y^2) + x(dy/dx) + 2y(dy/dx) = 0
2x + xy + y^2 + x(dy/dx) + 2y(dy/dx) = 0

Since we are looking for dy/dx = 0, we can simplify the equation further:
2x + xy + y^2 = 0

Now we have a system of equations:
x^2 + xy + y^2 = 7
2x + xy + y^2 = 0

By solving these two equations simultaneously, we can find the points where the tangent is parallel to the x-axis.

(b) Tangent parallel to the y-axis:
When the tangent is parallel to the y-axis, dy/dx is not defined, but dx/dy is.

Differentiating the equation implicitly with respect to y:
2x(dx/dy) + x(dy/dy) + 2y + y(dx/dy) + x = 0

Simplifying the equation:
2x(dx/dy) + x(dy/dy) + 2y + y(dx/dy) + x = 0

Since dy/dx is not defined, we solve for dx/dy. By setting dy/dy = 0 and solving the equation, we can find the value of dx/dy at these points.

To summarize, to find the points on the curve:
(a) Solve the system of equations 2x + xy + y^2 = 0 and x^2 + xy + y^2 = 7 to find the points where the tangent is parallel to the x-axis.
(b) Set dy/dy = 0 in the equation 2x(dx/dy) + x(dy/dy) + 2y + y(dx/dy) + x = 0, and solve for dx/dy to find its value at the points where the tangent is parallel to the y-axis.

To find the points on the curve where the tangent is parallel to the x-axis, we need to find the derivative of y with respect to x and set it equal to zero. This will give us the slope of the tangent line at any point on the curve.

Let's differentiate the given equation with respect to x:
d/dx (x^2 + xy + y^2) = d/dx (7)
2x + (x(dy/dx) + y) + 2y(dy/dx) = 0

Now, let's simplify the equation:
2x + xy' + y + 2yy' = 0

To find the points where the tangent is parallel to the x-axis, the slope dy/dx should be zero.

Setting dy/dx = 0 in the equation above:
2x + y = 0

This is a linear equation in terms of x and y. Solving for y, we get:
y = -2x

So, the points on the curve where the tangent is parallel to the x-axis are given by the equation y = -2x.

Moving on to finding the points on the curve where the tangent is parallel to the y-axis. In this case, we need to find the derivative of x with respect to y and set it equal to zero. This will give us the slope of the tangent line at any point on the curve.

Let's differentiate the equation with respect to y:
d/dy (x^2 + xy + y^2) = d/dy (7)
2x(dx/dy) + x + 2y = 0

Now, let's simplify the equation:
2x(dx/dy) + x + 2y = 0

To find the points where the tangent is parallel to the y-axis, the slope dx/dy should be zero.

Setting dx/dy = 0 in the equation above:
x + 2y = 0

This is another linear equation in terms of x and y. Solving for x, we get:
x = -2y

So, the points on the curve where the tangent is parallel to the y-axis are given by the equation x = -2y.

Finally, to find the value of dx/dy at the points where the tangent is parallel to the y-axis, we need to differentiate the equation x = -2y with respect to y.

Differentiating x = -2y with respect to y:
dx/dy = -2

Therefore, at the points where the tangent is parallel to the y-axis, dx/dy has a constant value of -2.