Chapter 2.3, Problem 41E

To determine

To calculate:The vertex, focus, directrix and the axis of symmetryof the given parabola and compare the given parabola with its standard equation with p =1 and vertex (0,0).

Explanation of Solution

Given information:The equation of parabola is $y=\frac{1}{8}{\left(x-3\right)}^2+2$ .

Formula used:

The parabola of the form ${\left(x-h\right)}^2=4p\left(y-k\right),\text{is}\text{a}\text{vertical}\text{parabola}$

The parabola of the form ${\left(y-k\right)}^2=4p\left(x-h\right),\text{is}\text{a}\text{horizontal}\text{parabola}$

Calculation:

The equation of given parabola can be written as

  $\begin{array}{l}\left(y-2\right)=\frac{1}{8}{\left(x-3\right)}^2\\ {\left(x-3\right)}^2=8\left(y-2\right)\mathrm{.....}\left(1\right)\end{array}$

Which is of the form ${\left(x-h\right)}^2=4p\left(y-k\right)$ .....(2), therefore it is vertical parabola.

By comparing equation (1) & (2)

  $\left(h,k\right)=\left(3,2\right)\&4p=8$ .

Therefore the vertex is $\left(h,k\right)=\left(3,2\right)$ .

Also,

  $\begin{array}{l}4p=8\\ p=2\end{array}$

Since p is positive that means the parabola opens upward.

Then focus is $\left(h,k+p\right)=\left(3,4\right)$ .

The directrix is at − p distance from the vertex and here the directrix is a horizontal line that passes from $\left(h,k-p\right)=\left(3,0\right)$ , therefore the equation of directrix is $y=0$ .

The axis of symmetry is the line that passes from the vertex $\left(h,k\right)=\left(3,2\right)$ and the focus $\left(h,k+p\right)=\left(3,4\right)$ .therefore the axis of symmetry is $x=3$ .

The standard equation of this parabola is ${\left(x-h\right)}^2=4p\left(y-k\right)$ , put p =1 and vertex (0,0)

Then the equation is $x^2=4y$ ........(3)

By comparing (1) and (3),it can be observed that

Vertex is shifted from $\left(0,0\right)\text{ to }(3,2)$

Focus is shifted from $\left(0,1\right)\text{to}\left(3,4\right)$

Directrix is shifted from $y=-1\text{to}y=0$

Axis of symmetry is shifted from $x=0\text{to}x=3$