Question

What is the focus of the parabola? y=14x2−x−1 Enter your answer in the boxes.

Answer 1

y=14x^2-x-1

If the directix line is given, the vertex is the midpoint between the focus and the directix.

in this case the answer is (1/28, -1)

Answer 2

Focus =
`((1)/(28), -1)`

Explanation:

The standard form is

`(x -h)^2 = 4p(y-k)`, ....[1] where the focus is (h, k + p).

Given the equation of parabola:

`y = 14x^2-x-1`

Using the square completing method.

Divide all the term by 14 on right side we have;

`y = 14(x^2-(1)/(14)x-(1)/(14))`

Now, complete the square on the right side.

`y =14(x^2-(1)/(14)x-(1)/(14)+((1)/(28))^2-((1)/(28))^2)`

then;

`y = 14((x-(1)/(28))^2-(1)/(784)-(1)/(14))`

⇒
`y = 14 \cdot ((x-(1)/(28))^2-(57)/(784))`

⇒
`y = 14(x-(1)/(28))^2-(57)/(56)`

then;

`y+(57)/(56) = 14 \cdot (x-(1)/(28))^2`

⇒
`(x-(1)/(28))^2 = (1)/(14)(y+(57)/(56))`

On comparing with [1] we have;

`h = (1)/(28)` ,
`k = -(57)/(56)` and
`4p = (1)/(14)`

⇒
`p = (1)/(56)`

`k+p = -(57)/(56)+(1)/(56) = (-56)/(56) = -1`

⇒Focus =
`((1)/(28), -1)`

Therefore, the focus of the parabola is,
`((1)/(28), -1)`