Question

finding the vertex, focus, directrix and focal width of y^2=16x

Answer 1

Write the parabola in vertex form:
\[y^2=16x~~\iff~~(y-0)^2=16(x-0)\]
The vertex is \((0,0)\).

Since \(y\) is begin squared, and has a positive coefficient, the plot of the parabola will "open" to the right, which means the focal point will be of the form \(F(x,0)\), where \(x\) is positive.

Any two points on the parabola will be equidistant from the focus. Take any two points, such as \((0,0)\) and \((1,4)\). Then,
\[\sqrt{(0-x)^2+(0-0)^2}=\sqrt{(1-x)^2+(4-0)^2}~~\iff~~x=\frac{17}{2}\]
So the focus is \(F\left(\dfrac{17}{2},0\right)\).

The focal width is the length of the chord drawn perpendicular to the axis of symmetry (which is \(y=0\)) that intersects the focus:

\[y^2=16\left(\frac{17}{2}\right)~~\iff~~y=\pm2\sqrt{34}~~\Rightarrow~~FW=4\sqrt{34}\]
The directrix is the reflection across the \(y\) axis of the line containing the focal width. Since the focal width lies on \(x=\dfrac{17}{2}\), the directrix is \(x=-\dfrac{17}{2}\).