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Chapter 6: Problem 24

Find the standard form of the equation of the parabola with the given characteristic(s) and vertex at the origin. Focus: (0,-2)

Short Answer

Expert verified
The standard form for the equation of the parabola with vertex at the origin and focus at (0, -2) is \(y = 0.5x^2\).

Step by step solution

01

Identify the Form of Equation

The vertex of the parabola is at the origin and the focus is on the y-axis. Since the parabola opens downward, we can say that it's a vertically aligned parabola. Hence, our equation will be in the form \( y = ax^2 \).
02

Determine the Value of 'a'

Our focus is at (0,-2) and the vertex is at the origin (0,0). The distance from the vertex to the focus is: \[ d = \sqrt{ (0-0)^2 + (-2-0)^2 } = 2 \] In the standard equation of a parabola \( y = ax^2 \), the distance from the vertex to the focus is represented by \( 4a \). Setting the distances equal to each other, we find: \[ 4a = 2 \] Solving for 'a', our constant value is: \[ a = 2 / 4 = 0.5 \]
03

Substitute 'a' into The Equation

Now that we found our 'a' value, we can substitute it into our equation. So we have our equation: \[ y = 0.5x^2 \]

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Most popular questions from this chapter

Water is flowing from a horizontal pipe 48 feet above the ground. The falling stream of water has the shape of a parabola whose vertex (0,48) is at the end of the pipe (see figure). The stream of water strikes the ground at the point \((10 \sqrt{3}, 0)\). Find the equation of the path taken by the water.

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