Find all points on the curve x^2y^2+ xy = 2 where the slope of the tangent is -1.    

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The slope of the tangent to a curve at any point is equal to the value of `dy/dx` at that point.

For the curve x^2y^2+ xy = 2

x^2*2y*y' + y^2*2x + xy' + y = 0

=> y' = -(x + x^2*2y)/(y + y^2*2x)

If the slope of the tangent is -1

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

=> x + x^2*2y = y + y^2*2x

=> x(1 + 2xy) = y(1 + 2xy)

=> x = y

As x^2y^2+ xy = 2

x^4 + x^2 = 2

=> x^4 + 2x^2 - x^2 - 2 = 0

=> x^2(x^2 + 2) - 1(x^2 + 2) = 0

=> x^2 = 1 and x^2 = -2

x = -2 can be eliminated as the roots are imaginary. This gives x = 1 and x = -1

At the points (1, 1) and (-1, -1) the slope of the tangent to the curve x^2y^2+ xy = 2 is -1.

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question