Question

What is the focus of the parabola? y=14x2−x−1

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Answer 1

This is a vertical parabola which opens upwards
we have the general formula
(x - h)^2 = 4p(y - k) where p is the y coordinate of the focus

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

add 1/14 to both sides

y/14 + 1 /14 = x^2 - x /14

now complete the square

y/14 + 1/14 = x^2 - x/14 + 1/28^2

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

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


comparing this with the general form:-
4p = 1/14
p = 1/56

so the focus is at (h,p) = (1/28, 1/56)

Answer 2

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

Explanation:

The given equation is

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

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

Add and subtract
`((-b)/(2))^2` in the equation to find the vertex form of the parabola.

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

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

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

The standard equation of parabola is

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

Where, (h,k+p) is focus.

It can be written as

`y=(1)/(4p)(x-h)^2+k` .... (2)

From (1) and (2), we get

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

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

`(h,k+p)=((1)/(28),-1)`

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