Find the focus and directrix of the parabola with the given equation. Then graph the parabola. 8x² + 4y = 0

Verified step by step guidance

1. Rewrite the given equation \$8x^2 + 4y = 0\( in a form that isolates \)y\(. Start by subtracting \)8x^2\( from both sides to get \)4y = -8x^2$.
2. Divide both sides of the equation by 4 to solve for \(y\): \(y = -2x^2\). This is now in the form \(y = ax^2\), which represents a vertical parabola.
3. Recall the standard form of a vertical parabola with vertex at the origin is \(y = \frac{1}{4p} x^2\), where \(p\) is the distance from the vertex to the focus (and also to the directrix).
4. Compare \(y = -2x^2\) with \(y = \frac{1}{4p} x^2\) to identify \(\frac{1}{4p} = -2\). Solve for \(p\) by setting \$4p = \frac{1}{-2}\(, which gives \)p = -\frac{1}{8}$.
5. Use the value of \(p\) to find the focus and directrix: since \(p\) is negative, the parabola opens downward. The focus is at \((0, p)\), and the directrix is the line \(y = -p\). Write these explicitly using the value of \(p\).

Key Concepts

Standard Form of a Parabola

A parabola can be expressed in standard form as either (x - h)^2 = 4p(y - k) for vertical parabolas or (y - k)^2 = 4p(x - h) for horizontal parabolas. Converting the given equation into one of these forms helps identify the vertex, focus, and directrix.

Focus and Directrix of a Parabola

The focus is a fixed point inside the parabola, and the directrix is a line outside it, both equidistant from any point on the parabola. The parameter p in the standard form determines the distance from the vertex to the focus and directrix.

Graphing a Parabola

Graphing involves plotting the vertex, focus, and directrix, then sketching the curve that is equidistant from the focus and directrix. Understanding the orientation (vertical or horizontal) and the width of the parabola is essential for an accurate graph.