Understanding the Parabola with Vertex at Origin and Axis Parallel to X-Axis

In mathematics, the parabola is a fundamental conic section. This article will guide you through the process of finding the equation of a parabola that has a vertex at the origin (0,0) and an axis parallel to the x-axis. We'll also cover the standard forms of such parabolas and how to determine the equation when a specific point lies on it.

Standard Form of a Parabola with the Vertex at the Origin and Axis Along the X-Axis

The standard form of the equation for a parabola with a vertex at the origin and an axis along the x-axis is:

y^2 4ax

In this equation:

a is a positive constant that represents the distance from the vertex to the focus. The focus is located at (a, 0). The directrix is the vertical line x -a.

Deriving the Equation from a Given Point

Suppose we need to find the equation of a parabola with the vertex at the origin and passing through a specific point, for example, (6, -2).

Step-by-Step Calculation

Step 1: Write the standard equation of the parabola with the vertex at the origin and axis along the x-axis:
y^2 4ax

Step 2: Substitute the given coordinates (6, -2) into the equation to find the value of a.

Substituting x 6 and y -2 into the equation:

(-2)^2 4a(6)

Simplifying, we get:

4 24a

Solving for a:

a frac{4}{24} frac{1}{6}

Step 3: Substitute a back into the standard equation to get the specific equation of the parabola:

y^2 4 times frac{1}{6} x

Simplifying, we get:

y^2 frac{2}{3} x

Additional Examples and Variations

Let's consider a few additional scenarios:

Example 1: Passing Through (0, -3) and (6, -2)

Given the parabola passes through both (0, -3) and (6, -2), we can use the equation of the parabola with the vertex at the origin:

y^2 4ax

Substituting (6, -2):

(-2)^2 4a(6) Rightarrow 4 24a Rightarrow a frac{4}{24} frac{1}{6}

Thus, the equation is:

y^2 frac{2}{3} x

Example 2: Passing Through a Point with a Negative a

Consider a similar scenario where the parabola passes through (0, -3) and (6, -2), but this time the directrix is parallel to the x-axis. The equation will be:

x a y^2

Substituting (0, -3):

0 a (-3^2) Rightarrow 0 9a Rightarrow a -frac{1}{12}

Thus, the equation is:

x -frac{1}{12} y^2

Example 3: Finding the Equation Using a Conic Section

For a parabola with vertex at (0, 0) and the axis along the x-axis, the general form is:

y^2 4ax

Given the point (6, -2), we substitute:

(-2)^2 4a(6) Rightarrow 4 24a Rightarrow a frac{1}{6}

So, the equation is:

y^2 frac{2}{3} x

Conclusion

Understanding how to find the equation of a parabola with the vertex at the origin and axis along the x-axis is essential in geometry and algebra. By using the standard form and substituting specific points, we can derive the unique equation. This process also helps us in visualizing the geometric properties of parabolas.