Chapter 11.4, Problem 2E

The graphs of x² = 12y and (x − 3)² = 12(y − 1) are given. Label the focus, directrix, and vertex on each parabola.

To determine

To label: The focus, directrix and vertex on each of the parabolas $x^2=12y$ and ${\left(x-3\right)}^2=12\left(y-1\right)$.

Answer to Problem 2E

From the equation of parabola, $x^2=12y$, the vertex is $V\left(0,0\right)$, the focus is $F\left(0,3\right)$ and the directrix is $y=-3$.

From the equation of parabola, ${\left(x-3\right)}^2=12\left(y-1\right)$, the vertex is $V\left(3,1\right)$, the focus is $F\left(3,4\right)$ and the directrix is $y=-2$.

Explanation of Solution

Definition used:

Definition 1:

The equation of the parabola with vertex $V(0,0)$, focus $F(0,p)$ and directrix $y=-p$ is $x^2=4py$.

Definition 2:

If $h$ and $k$ are any two positive real numbers and a graph of any equation is in terms of $x$ and $y$, then

(a) When replace $x$ by $x-h$, the graph shifts right by $h$ units.

(b) When replace $x$ by $x+h$, the graph shifts left by $h$ units.

(c) When replace $y$ by $y-k$, the graph shifts upward by $k$ units.

(d) When replace $y$ by $y+k$, the graph shifts downward by $k$ units.

Calculation:

Compare the parabola $x^2=12y$ to the general form of parabola $x^2=4py$.

Therefore,

$\begin{array}{l}4p=12\\ p=\frac{12}{4}\\ p=3\end{array}$

Replace $x$ by $x-3$ and $y$ by $y-1$ in the equation $x^2=12y$ leads to ${\left(x-3\right)}^2=12\left(y-1\right)$.

From the equation of parabola, $x^2=12y$, obtain the values of vertex, focus and directrix.

The vertex is $V\left(0,0\right)$, the focus is $F\left(0,3\right)$ and the directrix is $y=-3$.

From the equation of parabola ${\left(x-3\right)}^2=12\left(y-1\right)$, obtain the values of the vertex, focus and directrix.

The sum $0+3=3$ results in the x-coordinate and the sum $0+1=1$ results in the y-coordinate of the vertex of the shifted parabola, that is, the vertex becomes $V\left(3,1\right)$.

The sum $0+3=3$ results in the x-coordinate and the sum $3+1=4$ results in the y-coordinate of the focus of the shifted parabola, that is, the focus becomes $F\left(3,4\right)$

The sum $-3+1=-2$ results in the shift of the line $y=-3$ to $y=-2$.

Therefore, for the equation of parabola, $x^2=12y$, the vertex is $V\left(0,0\right)$, the focus is $F\left(0,3\right)$, and the directrix is $y=-3$, and for the equation of parabola, ${\left(x-3\right)}^2=12\left(y-1\right)$, the vertex is $V\left(3,1\right)$, the focus is $F\left(3,4\right)$, and the directrix is $y=-2$.