Let's solve the given problem.

Problem:

We are asked to find the equation of a parabola with the following properties:

• Vertex: $(0, 0)$
• Axis along the x-axis
• Passes through the point $(2, -1)$

Solution:

Since the axis of the parabola is along the x-axis, its equation will be in the form:

$y^2 = 4ax$

where:

• $a$ is a constant that determines how wide or narrow the parabola is.
• The vertex is at the origin $(0, 0)$, and the axis of symmetry is the x-axis.

Step 1: Use the given point $(2, -1)$

We are given that the parabola passes through $(2, -1)$. This means the point must satisfy the equation $y^2 = 4ax$.

• For $x = 2$ and $y = -1$, substitute these values into the equation:

$(-1)^2 = 4a(2)$

$1 = 8a$

Step 2: Solve for $a$

$a = \frac{1}{8}$

Step 3: Write the Equation of the Parabola

Now that we know $a = \frac{1}{8}$, the equation of the parabola is:

$y^2 = 4 \times \frac{1}{8} \times x$

$y^2 = \frac{1}{2}x$

Final Equation:

$y^2 = \frac{1}{2}x$

Sketch (Description):

• Vertex: The vertex is at $(0, 0)$.
• Orientation: Since the equation is $y^2 = \frac{1}{2}x$, the parabola opens rightward along the positive x-axis.
• Point on the parabola: The parabola passes through the point $(2, -1)$.

Do you need a plot of the parabola or further clarifications?

Here are 5 related questions you could explore:

1. What happens if the axis of the parabola is along the y-axis instead?
2. How do you derive the general form of a parabola given any vertex?
3. How would the equation change if the parabola were shifted to another vertex, say $(h, k)$?
4. What are the focus and directrix of the parabola with this equation?
5. How would you convert this equation into parametric form?

Tip: Always substitute given points into your equation to verify solutions. This ensures accuracy when deriving parameters.