Chapter 2.3, Problem 43E

To determine

To calculate:The vertex, focus, directrix and the axis of symmetryof the given parabolaand 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 $x=\frac{1}{16}{\left(y-3\right)}^2+1$ .

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

  $\begin{array}{l}\left(x-1\right)=\frac{1}{16}{\left(y-3\right)}^2\\ {\left(y-3\right)}^2=16\left(x-1\right)\mathrm{.....}\left(1\right)\end{array}$

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

By comparing equation (1) & (2)

  $\left(h,k\right)=\left(1,3\right)\&4p=16$ .

Therefore the vertex is $\left(h,k\right)=\left(1,3\right)$ .

Also,

  $\begin{array}{l}4p=16\\ p=4\end{array}$

Since p is positive that means the parabola opens right.

Then focus is $\left(h+p,k\right)=\left(5,3\right)$ .

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

The axis of symmetry is the line that passes from the vertex $\left(h,k\right)=\left(1,3\right)$ and the focus $\left(h+p,k\right)=\left(5,3\right)$ .therefore the axis of symmetry is $y=3$ .

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

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

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

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

Focus is shifted from $\left(1,0\right)\text{to}\left(5,3\right)$

Directrix is shifted from $x=-1\text{to}x=-3$

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