Equation of Parabola with Curvature 4 at Origin: Need Help!

[fresh]

So i have this question which seems easy enough but maybe iam not thinking in the right mind frame or something.

The question is, find the equation of a parobola which has a curvature of 4 at the origin.

Some sort of hint/push in the right direction would be appreciated. Thanks

 

[Tide]

If the slope is zero at the origin then the curvature is just the second derivative of the function.

 

[M98Ranger]

Hi there,

Finding the equation of a parabola with a given curvature can be a bit tricky, but with the right approach, it can be solved. Here are some steps that might help you:

1. Recall that the curvature of a parabola is given by the formula: k = 1/2a, where a is the coefficient of the squared term in the equation.

2. Since the problem specifies that the curvature is 4 at the origin, we can set k = 4 and solve for a. This will give us the coefficient of the squared term in the equation.

3. Once we have a, we can plug it into the general equation of a parabola: y = ax^2 + bx + c.

4. Since the parabola is passing through the origin, we know that the y-intercept (c) is 0. This will simplify our equation to y = ax^2 + bx.

5. Now we need to find the value of b. To do this, we can use the fact that the curvature is also equal to the derivative of the equation. This means that we can take the derivative of y = ax^2 + bx and set it equal to k (which in this case is 4). This will give us an equation with b as the only unknown variable, which we can solve for.

6. Once we have the values of a and b, we can plug them back into the equation y = ax^2 + bx to get the final equation of the parabola.

I hope this helps and gives you a better understanding of how to approach this type of problem. Good luck!