Intersection and Orthogonality of Curves y24x and y2-2x26

This article explores the intersection points and the orthogonality of two algebraic curves defined by the equations y24x and y2-2x26. These curves are analyzed to determine their intersection points, and the orthogonality of the curves at these points is verified.

Intersection Points

To find the intersection points of the curves, we start with the given equations:

y24x y2-2x26

First, we substitute the expression for y2 from equation (1) into equation (2):

y2-2x26 becomes

4x-2x26

Arranging the equation, we get:

-2x2 4x - 6 0

Multiplying the equation by -1 to simplify:

2x2 - 4x 6 0

Further simplifying:

x2 - 2x 3 0

Factoring the quadratic equation:

(x - 3)(x 1) 0

This gives us the solutions:

x 1, x -3

Now, we substitute these values of x back into the equation y24x to find the corresponding y-values.

x 1

y24(1)

y24

y ±2

So, the intersection points are (1, 2) and (1, -2).

x -3

y24(-3)

y2 -12

This equation has no real solutions, but we can consider the complex solutions:

y ±i√12

So, the intersection points are (-3, i√12) and (-3, -i√12).

Thus, the only real intersection points are (1, 2) and (1, -2).

Orthogonality of the Curves

To determine if the curves are orthogonal at the intersection points, we first find the derivatives of the respective equations to find the slopes of the tangents:

Derivative of y24x

2y(dy/dx)4

dy/dx2/y

Derivative of y2-2x26

2y(dy/dx)-4x0

dy/dx2x/y

Checking the Intersection Points

(1, 2)

dy/dx for y24x at (1, 2)

dy/dx2/21

dy/dx for y2-2x26 at (1, 2)

dy/dx2(1)/21

The product of these derivatives is:

1 * 1 1

This is not a negative reciprocal, which means the curves are not orthogonal at this point.

(1, -2)

dy/dx for y24x at (1, -2)

dy/dx2/(-2)-1

dy/dx for y2-2x26 at (1, -2)

dy/dx-2(1)/(-2)1

The product of these derivatives is:

-1 * 1 -1

This is a negative reciprocal, which means the curves are orthogonal at this point.

Complex Points

Considering the complex points (-3, i√12) and (-3, -i√12), the method remains the same. However, the derivatives at these points do not provide real-valued slopes, which makes the concept of orthogonality in the complex plane less straightforward.

Conclusion

The only real intersection points of the curves y24x and y2-2x26 are (1, 2) and (1, -2). The curves are orthogonal at the point (1, -2).

Link to further reading on algebraic curves and their properties.