Question

What is the focus of the parabola? y=14x2−x−1 Enter your answer in the space.. ( ?,? )

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

Answer:

The focus of the parabola is $(\frac{1}{28},-1)$.

Step-by-step explanation:

The given equation is

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

$y=14(x^2-\frac{x}{14})-1$

Add and subtract $(\frac{-b}{2})^2$ in the equation to find the vertex form of the parabola.

$y=14(x^2-\frac{x}{14}+(\frac{1}{28})^2-(\frac{1}{28})^2)-1$

$y=14(x-\frac{1}{28})^2-\frac{1}{56}-1$

$y=14(x-\frac{1}{28})^2-\frac{57}{56}$                ... (1)

The standard equation of parabola is

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

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

It can be written as

$y=\frac{1}{4p}(x-h)^2+k$                               .... (2)

From (1) and (2), we get

$h=\frac{1}{28},k=\frac{-57}{56},\frac{1}{4p}=14\Rightarrow p=\frac{1}{56}$

$k+p=\frac{-57}{56}+\frac{1}{56}=\frac{-56}{56}=-1$

$(h,k+p)=(\frac{1}{28},-1)$

Therefore the focus of the parabola is $(\frac{1}{28},-1)$.