Help with a Cal problem plz?

Question

How can I show that the curves 3x^2+2x-3y^2=1 and 6xy+2y=0 are orthogonal?

Answer 1

I assume you mean that the curves are orthogonal where they meet.

First find the point(s) of intersection.

6xy + 2y = 0
3xy + y = 0
(3x + 1)y = 0

x = -1/3 or y = 0

Plug these into the first equation and solve for the other variable.

For x = -1/3, solve for y.

3x² + 2x - 3y² = 1
3(-1/3)² + 2(-1/3) - 3y² = 1
3/9 - 2/3 - 3y² = 1
-1/3 - 3y² = 1
-3y² = 4/3
y² = -4/9

No solution.

For y = 0, solve for x.

3x² + 2x - 3*0 = 1
3x² + 2x - 1 = 0
(3x - 1)(x + 1) = 0
x = 1/3, -1

Our points of solution are (1/3, 0) and (-1, 0).
______________

Now take the derivatives of the curves implicitly at solve for the slopes at the points of intersection.

For the first curve:

6xy + 2y = 0
3xy + y = 0
3y + 3x(dy/dx) + 1(dy/dx) = 0
(3x + 1)(dy/dx) = -3y
dy/dx = -3y/(3x + 1)

dy/dx = 0 at both (1/3, 0) and (-1, 0)

For the second curve:

3x² + 2x - 3y² = 1
6x + 2 - 6y(dy/dx) = 0
-6y(dy/dx) = -(6x + 2)
dy/dx = (6x + 2)/(6y) = (3x + 1)/(3y)

dy/dx is undefined at both (1/3, 0) and (-1, 0).

The first curve in horizontal at their points of intersection and the second curve is vertical. Therefore the curves are orthogonal at their points of intersection.