Question

Find the vertex, focus, directrix, and focal width of the parabola. x = 4y2.

Answer 1

The vertex is (0,0), focus of the parabola is
`((1)/(16),0)`, directrix of the parabola is
`y=-(1)/(16)`, focal width is
`(1)/(4)`.

Explanation:

The given equation of parabola is

`x=4y^2`

It can be written as

`y^2=(1)/(4)x` ....(1)

The general equation of parabola is

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

Where, (h,k) is vertex, (h+p,k) is focus, y=h-p is directrix and |4p| is focal width.

On comparing (1) and (2), we get

`h=0,k=0`

The vertex is (0,0).

`4p=(1)/(4)`

`p=(1)/(16)`

Focus of the parabola is

`(h+p,k)=(0+(1)/(16),0)=((1)/(16),0)`

Therefore focus of the parabola is
`((1)/(16),0)`.

Directrix of the parabola is

`y=h-p=0-(1)/(16)=-(1)/(16)`

Directrix of the parabola is
`y=-(1)/(16)`.

Focal width is

`|4p|=|4* (1)/(16)|=(1)/(4)`

Focal width is
`(1)/(4)`.

Answer 2

- Vertex : ( 0, 0 ).
- Focus: ( p/2, 0 );
y² = 1/4 x
y² = 2 p x ⇒ 2 p = 1/4
p = 1/8
p/2 = 1/16
F ( 1/16, 0)
- Directrix:
x = - p/2 x
x = - 1/16 x
- The focal length:
2 p = 2 · 1/8 = 1/4