what is the function for a parabola with a starting angle?

Question

lets say i want a parabola that starts out at a 35 degree angle. what would be its formula?

please explain how you solved this because i want to be able to solve for the formula of every other angle as well.

Answer 1

Well, I don't know what you mean by the parabola "starting" at a certain angle, because the center of the parabola is always going to be flat if this is a single-valued function. However, it sounds like you want a way to adjust a parabola so that at a certain point its tangent line is pointed with direction angle theta.

Just set the derivative at that point to tan(theta).

For a parabola y = ax^2 + bx + c, its derivative at x is 2a * x + b. So you just set 2a * x + b = tan(theta). You can choose whatever values you want for two of the three variables on the left-hand side.

Answer 2

The general form of a quadratic equation is:

ax² + bxy + cy² + dx + ey + f = 0

The presence of a non-zero xy term indicates that the curve is rotated and not directly aligned with either the x or y axis.

Let's pick the following equation.

x² - 4xy + 4y² + 2x + 16y + 15 = 0

The discriminant tells us what kind of a curve it is. The discriminant is

b² - 4ac

If it is a second order equation and the discriminant equals zero, then the equation is that of a parabola. Here we have:

b² - 4ac = 4² - 4*1*4 = 16 - 16 = 0

So it is a parabola. To find its angle of rotation θ, take the formula:

cot(2θ) = (a - c) / b = (1 - 4) / -4 = 3/4

2θ = arccot(3/4)
θ = (1/2)arccot(3/4) ≈ 26.565051°