Post 1
The equation y = cx2 has been solved by “manual fitting” of the figures. However, this is not the appropriate way, since it is not universal, and it does not suit for all the problems. Instead, x and y values (x = -2, y = 6 and x = 2, y = 6) should be applied to the equation. For example, 6 = 22∙ c; 4 ∙ c = 6, subtracting by 4, c = 1.5. This method is universal and can be used for any other complicated case to find the equation. Then, the focus can be determined.
If the algebraic method to find the parabola equation is applied, the checking operation is redundant since the solution is straightforward. It is very important to apply simple and mathematically justified methods for solution of the real-world problems. Sometimes, the problems are complicated, and their solution requires training. The training is performed with the simple problems, as the one discussed in this post.
Post 2
The problem does not suggest any coordinates for the vertex of the parabola. Therefore, the choice of (0, 0) vertex is fully justified. However, in this case the equation:
x2 = 4 ∙ p ∙ y
would have been reasonable to use instead of (x - h)2 = 4 ∙ p ∙ (y – k).
The proposed simplified equation is the special case of the vertex equation. Although the proposed equation and the used in the problem provide the same solution, the application of geometry and mathematics for the real-world problems has to use the simplest forms of the equations.
Actually, the problem with an equation for vertex with (0, 0) coordinates has to be the first in the course for parabols application in real-life problems as it is simple and easy to solve.
