Problem 27 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 19-30: Find an equation of the parabola that satisfies the given conditions. Vertex at the origin, symmetric to the \(y\)-axis, and passing through the point \((2,-3)\)

Step-by-Step Solution

Answer

The equation of the parabola is \( y = \frac{-3}{4}x^2 \).

Step 1: Determine the Basic Parabola Equation

Since the parabola is symmetric to the y-axis and its vertex is at the origin, the basic equation of the parabola is given by \( y = ax^2 \). We need to determine the value of \( a \).

Step 2: Use the Given Point to Find 'a'

Substitute the coordinates of the given point \((2, -3)\) into the equation \( y = ax^2 \). This gives \( -3 = a(2)^2 \).

Step 3: Solve for 'a'

Simplify the equation from Step 2 to find \( a \). \( -3 = 4a \) implies \( a = \frac{-3}{4} \).

Step 4: Write the Final Equation of the Parabola

Substitute the value of \( a \) back into the basic equation: \( y = ax^2 \). Thus, the final equation of the parabola is \( y = \frac{-3}{4}x^2 \).

Key Concepts

Symmetric About the Y-axis

A parabola that is symmetric about the y-axis is one that looks the same on both sides of the y-axis, like a mirror image. This type of symmetry is due to a special property of the parabola's equation: it contains only even powers of the variable in the equation.
For this exercise, our parabola has its vertex at the origin and is symmetric about the y-axis. Therefore, the standard form of the equation will be:

• Basic form: \( y = ax^2 \)
• Notice: There are no terms with odd powers of \( x \) or any \( y \)-terms aside from the leading term itself.

This symmetry ensures that if a point \((x, y)\) lies on the parabola, then \((-x, y)\) will also lie on it. Thus, the equation \( y = ax^2 \) accurately reflects the symmetry about the y-axis.

Vertex at the Origin

When a parabola's vertex is at the origin, it means the peak or the lowest point, depending on the direction the parabola opens, is at the point \((0,0)\). The vertex is a significant component of a parabola because it aids in determining its shape and position on the graph.
Given the conditions that the parabola is symmetric about the y-axis and the vertex lies at the origin, the parabola takes the form \( y = ax^2 \). This ensures:

• The parabola opens either upwards or downwards.
• It will always pass through the origin \((0,0)\).

The combination of these parameters simplifies graphing and understanding parabolas, especially those symmetric around the y-axis.

Using a Given Point

To find the specific equation of a parabola, sometimes you need more than just its symmetry and vertex information. You often use a given point that the parabola passes through to determine the coefficient, \( a \), in the equation.
In this exercise, we're given the point \((2, -3)\). By substituting this point into our basic equation \( y = ax^2 \), we get:

• \(-3 = a(2)^2\)
• Which simplifies to \(-3 = 4a\)
• Solving for \( a \), we find \( a = \frac{-3}{4} \)

Incorporating this value back into our equation gives us the specific parabolic equation \( y = \frac{-3}{4}x^2 \). This equation now precisely describes our parabola's shape, direction, and alignment.