Chapter 2.3, Problem 42E

To determine

To calculate: The vertex, focus, directrix and the axis of symmetry of 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 $y=-\frac{1}{4}{\left(x+2\right)}^2+1$ .

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

  $\begin{array}{l}\left(y-1\right)=-\frac{1}{4}{\left(x+2\right)}^2\\ {\left(x+2\right)}^2=-4\left(y-1\right)\mathrm{.....}\left(1\right)\end{array}$

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)

  $\left(h,k\right)=\left(-2,1\right)\&4p=-4$ .

Therefore the vertex is $\left(h,k\right)=\left(-2,1\right)$ .

Also,

  $\begin{array}{l}4p=-4\\ p=-1\end{array}$

Since p is negative that means the parabola opens downward.

Then focus is $\left(h,k+p\right)=\left(-2,0\right)$ .

The directrix is at − p distance from the vertex and here the directrix is a horizontal line that passes from $\left(h,k-p\right)=\left(-2,2\right)$ , therefore the equation of directrix is $y=2$ .

The axis of symmetry is the line that passes from the vertex $\left(h,k\right)=\left(-2,1\right)$ and the focus $\left(h,k+p\right)=\left(-2,0\right)$ .therefore the axis of symmetry is $x=-2$ .

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 $\left(0,0\right)\text{ to }(-2,1)$

Focus is shifted from $\left(0,1\right)\text{to}\left(-2,0\right)$

Directrix is shifted from $y=-1\text{to}y=2$

Axis of symmetry is shifted from $x=0\text{to}x=-2$