`y^2=28x` Graph the equation. Identify the focus, directrix, and axis of symmetry of the parabola.

Expert Answers
lemjay eNotes educator| Certified Educator

`y^2=28x`

Take note that one of the vertex form of parabola is

`(y - k)^2 = 4p(x-h)`

where

(h,k) is the vertex and,

p is the distance between vertex and focus and also the same distance between the vertex and directrix.

So to graph it, first find the vertex.

Rewriting the given equation in exact form as above, it becomes:

`(y-0)^2 =28(x-0)`

So the vertex of the parabola is (0,0).

Then, determine the points of the parabola. To do so, isolate the y.

`y^2=28x`

`y=+-sqrt(28x)`

`y=+-2sqrt(7x)`

Then, assign a value to x. And solve for the y values.

`x=7,y=+-2sqrt(7*7) =+-2*7=+-14`

Plot these three points (0,0), (7,-14) and (7,14). And, connect them.

Therefore, the graph of the given equation is:

To determine the focus and directrix, consider the coefficient of the unsquared portion of the given equation and set it equal to 4p.

`4p=28`

And solve for p.

`p=28/4`

`p=7`

So both the focus and the directrix are 7 units from the vertex.

Since the parabola opens to the right, the coordinates of the focus is:

`(h + p, k) = (0+7, 0) = (7,0)`

And the equation of directrix is:

`x=h-k`

`x=0-7`

`x=-7`

Therefore, the focus is `(7,0)` and the directrix is `x=-7` .

Take note that the axis of symmetry of a parabola is a line that passes the vertex and the focus. And it is perpendicular to the directrix. (See attachment.)

Therefore, the equation of its axis of symmetry is `y=0` .

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