Chapter 11, Problem 3RE

a.

To determine

To find: The vertex, focus and directrix of the given parabola.

Answer to Problem 3RE

  $\begin{array}{l}\text{Vertex}=\left(0,0\right)\\ \text{Focus}=\left(0,-2\right)\\ \text{Directrix}\Rightarrow y\text{=2}\end{array}$

Explanation of Solution

Given information:

  $x^2+8y=0$

Calculation:

  ${\left(x-h\right)}^2=4p\left(y-k\right)$

  $\begin{array}{l}\text{Given equation is }{x}^2+8y=0\\ x^2=-8y\mathrm{.......}\left(1\right)\end{array}$

Comparing (1) with standard formula. We get

  $\begin{array}{l}h=0,K=0,4p=-8\\ \therefore p=-2\\ \text{The vertex of parabola is}\left(h,K\right)=\left(0,0\right)\end{array}$

  $\begin{array}{l}\text{Focus of parabola is}=\left(h,K+P\right)\text{=}\left(0,0+\left(-2\right)\right)\\ =\left(0,-2\right)\end{array}$

  $\begin{array}{l}\text{Directrix of parabola is }y=K-p\\ y=0-\left(-2\right)\\ \therefore y=2\text{ is directrix}\end{array}$

This parabola is downward and is symmetry about y-axis.

Hence,

  $\begin{array}{l}\text{Vertex}=\left(0,0\right)\\ \text{Focus}=\left(0,-2\right)\\ \text{Directrix}\Rightarrow y\text{=2}\end{array}$

b.

To determine

To describe: How the graph of the function can be drawn.

Explanation of Solution

Given information:

Given equation $x^2+8y=0\text{ i}{\text{.e x}}^2=-8y$

Graph:

Given equation $x^2+8y=0\text{ i}{\text{.e x}}^2=-8y$

For $\begin{array}{l}y=0,x=0\\ y=-2,x=\pm 4\\ y=-8,x=\pm 8\\ \therefore \text{ The points graph would be }\left(0,0\right),\left(-2,4\right),\left(-2,-4\right),\left(-8,8\right),\left(-8,-8\right)\end{array}$

Using these points graph can be plotted.

  

Interpretation:

The graph of equation $x^2=-8y$ or $x^2+8y=0$ can be obtained by taking negative values of y.