2. 10 points Find all the points on the curve x 2 y 2 +xy = 2 where the slope of the tangent line to the curve is equal to −1.

Question

2. 10 points Find all the points on the curve x 2 y 2 +xy = 2 where the slope of the tangent line to the curve is equal to −1.

Answer 1

On the curve x²y² + xy = 2, the two points where the slope of the tangent line is equal to -1 is (1, 1) and (-1, -1).

We can use implicit differentiation to answer this question. First, implicitly differentiate both sides:

2xy² + 2x²yy' + y + xy' = 0

Then, isolate for y' to find the first derivative of the curve:

2x²yy' + xy' = -2xy² - y

y'(2x²y+x) = -2xy² - y

y' = (-2xy² - y)/(2x²y + x)

Then, plug in -1 to y' since we want to find the points where the slope of the tangent line is -1:

-1 = (-2xy² - y)/(2x²y + x)

Multiply both sides by (2xy + x):

-2x²y - x = -2xy² - y

Notice that the two sides will be equal if x = y. So, if we find the points that satisfy x = y and the original equation x²y² + xy = 2, we will find the points where the slope is -1. We can then substitute x = y into the original equation to isolate for x:

x⁴ + x² = 2

x⁴ + x² - 2 = 0

We can use a u-substitution to turn this 4th-degree polynomial into a quadratic:

u = x²

u² + u - 2 = 0

Which can then be factored:

(u + 2)(u - 1) = 0

u = -2, 1

u = -2 cannot be an answer since x² ≠ -2 so that leaves us with u = 1 or x² = 1. This gives us an answer of:

x = 1 and x = -1. Substituting into the original equation gives us the points (1, 1) and (-1, -1).

Answer 2

x²y² + xy = 2; 2xy² + 2x²yy' + y + xy' = 0. If y' = - 1 then 2xy² - 2x²y + y - x = 0; 2xy(y - x) + (y - x) = 0;

(y - x)(2xy + 1) = 0; y - x = 0; y = x; x⁴ + x² = 2; (x⁴ - 1) + (x² - 1) = 0; (x² - 1)(x² + 2) = 0; x² - 1 = 0; x = ±1;

x = 1, y = 1; x = - 1, y = - 1. So, (1, 1) and (-1, - 1);

2xy = - 1; xy = - 1/2; 1/4 - 1/2 = 2. So, this contradiction.

Answer (1, 1) and (-1, - 1)