Answer:
Focus = [tex](\frac{1}{28}, -1)[/tex]
Step-by-step explanation:
The standard form is
[tex](x -h)^2 = 4p(y-k)[/tex], ....[1] where the focus is (h, k + p).
Given the equation of parabola:
[tex]y = 14x^2-x-1[/tex]
Using the square completing method.
Divide all the term by 14 on right side we have;
[tex]y = 14(x^2-\frac{1}{14}x-\frac{1}{14})[/tex]
Now, complete the square on the right side.
[tex]y =14(x^2-\frac{1}{14}x-\frac{1}{14}+(\frac{1}{28})^2-(\frac{1}{28})^2)[/tex]
then;
[tex]y = 14((x-\frac{1}{28})^2-\frac{1}{784}-\frac{1}{14})[/tex]
⇒[tex]y = 14 \cdot ((x-\frac{1}{28})^2-\frac{57}{784})[/tex]
⇒[tex]y = 14(x-\frac{1}{28})^2-\frac{57}{56}[/tex]
then;
[tex]y+\frac{57}{56} = 14 \cdot (x-\frac{1}{28})^2[/tex]
⇒[tex](x-\frac{1}{28})^2 = \frac{1}{14}(y+\frac{57}{56})[/tex]
On comparing with [1] we have;
[tex]h = \frac{1}{28}[/tex] , [tex]k = -\frac{57}{56}[/tex] and [tex]4p = \frac{1}{14}[/tex]
⇒[tex]p = \frac{1}{56}[/tex]
[tex]k+p = -\frac{57}{56}+\frac{1}{56} = \frac{-56}{56} = -1[/tex]
⇒Focus = [tex](\frac{1}{28}, -1)[/tex]
Therefore, the focus of the parabola is, [tex](\frac{1}{28}, -1)[/tex]
