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Chapter 6: Problem 33

Find the vertex, focus, and directrix of the parabola. Then sketch the parabola. $$x^{2}+6 y=0$$

Short Answer

Expert verified
The vertex of the given parabola is at the origin (0,0), the focus is (0, -1.5) and the equation of the directrix is y = 1.5. Upon sketching the parabola, it will open downward towards the negative y-axis.

Step by step solution

01

Identify the Value of 'a'

Rearrange the given formula \(x^{2} +6y = 0\) which forms \(x^{2} = -6y\). Compared to the standard form \(x^{2}=4ay\), '4a' is equal to '-6'. Solving for 'a' gives \(a = \frac{-6}{4}= -1.5\)
02

Finding the Vertex

The vertex of the parabola is given by (h, k), where 'h' and 'k' are the coefficients of 'x' and 'y' respectively. In this case, there are no coefficients other than 1, thus the vertex of the parabola is at the origin, (0, 0).
03

Finding the Focus

Using the equation for the focus of a parabola, (h, k + a), where 'a' has been found -1.5. Substituting the values in, the focus of the parabola is (0, -1.5)
04

Finding the Directrix

The equation for the directrix of a parabola is y = k - a. Substituting the known values gives y = 0 -(-1.5), which simplifies to y = 1.5.
05

Sketching the Parabola

To draw the sketch, plot the vertex at (0,0) point. Because the parabola opens downward (since 'a' is negative), the focus is at (0, -1.5). Draw a horizontal line at y = 1.5 which is the directrix. Now, sketch the parabola, which opens downward and crosses the vertex, moving closer to the directrix as it extends but never touching it.

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