﻿ Equations of Simple Parabolas (Part 2)

# Equations of Simple Parabolas (Part 2)

## Shifting the Parabola

Now, here's some very good news. By using graphical transformations, knowledge of this one simple equation $\,y = ax^2\,$ actually gives full understanding of all parabolas with directrix parallel to the $x$-axis! The results are summarized next.

Shift the parabola (together with its focus and directrix) horizontally by $\,h\,,$ and vertically by $\,k\,.$ This yields the following information:

original equation:  $y=ax^2$
shifted equation:  $y=a(x-h)^2 + k$
original vertex:  $(0,0)$
new vertex:  $(h,k)$
original focus:  $\displaystyle (0,p) = (0,\frac{1}{4a})$
new focus:  $\displaystyle (h,p+k)= (h,\frac{1}{4a}+k)$
original directrix:  $y=-p$
new directrix:  $y=-p+k$

## Example(Graphing a Shifted Parabola)

In this example, I illustrate the approach that I usually take when asked to give complete information about the parabola $\,y = a(x-h)^2 + k\,.$

Question:  Completely describe the graph of the equation $\,y = -3(x+5)^2 + 1\,.$
Solution:
• Concave up or down?  Since $\,a=-3\lt 0\,,$ the parabola is concave down (sheds water). Since the focus is always inside a parabola, we know the focus is under the vertex.

• Find the vertex:  The vertex of  $\,y = a(x-h)^2 + k\,$  is the point $\,(h,k)\,$:

$$\cssId{s34}{y = a( \overset{\text{What value of } \ x\ }{\overset{\text{ makes this zero?}}{\overbrace{ \underset{\text{answer: } \ h}{\underbrace{x-h}}}}})^2\ \ \ + \overset{y\text{-value of the vertex}}{\overbrace{\ \ k\ \ }}}$$

For us, what makes $\,x+5\,$ equal to zero? Answer: $\,-5\,$

Thus, $\,-5\,$ is the $x$-value of the vertex. When $\,x = -5\,,$ the corresponding $y$-value is $\,1\,.$ Thus, the vertex is $\,(-5,1)\,.$

• Find the distance from the vertex to the focus: Using the ‘one over four pee pair’ memory device:

$$\cssId{s42}{p = \frac{1}{4a} = \frac{1}{4(-3)} = -\frac{1}{12}}$$

Thus, the distance from the focus to the vertex is:

$$\cssId{s44}{|p| = \left|-\frac1{12}\right| = \frac{1}{12}}$$
• Find the focus:  We already determined that the focus lies below the vertex. So, the focus is:

$$\cssId{s48}{(-5,1-\frac{1}{12}) =(-5,\frac{11}{12})}$$
• Find the equation of the directrix: Every horizontal line has an equation of the form:

$$\cssId{s51}{y = \text{(some #)}}$$

In our example, the focus lies $\,\frac{1}{12}\,$ below the vertex, so the directrix lies $\,\frac{1}{12}\,$ above the vertex. Thus, the equation of the directrix is $\,y=1+\frac{1}{12}\,,$ that is, $\,y=\frac{13}{12}\,.$

• Plot an additional point: Plotting an additional point gives a sense of the ‘width’ of the parabola. For example, when $\,x = -4\,$ we have:

\begin{align} y &= -3(-4+5)^2 + 1\cr &= -3(1) + 1\cr &= -2 \end{align}

In this parabola, the vertex is close to the focus, so the parabola is narrow.

$\,y=-3(x+5)^2+1\,$

For fun, zip up to WolframAlpha and type in any of the following:

vertex of y = -3(x+5)^2 + 1

focus of y = -3(x+5)^2 + 1

directrix of y = -3(x+5)^2 + 1

How easy is that?!