Solution:
The equation of a parabola is given as
[tex] y=14x^2-3x+18 [/tex]
we have been asked to find the coordinates of the focus of the parabola?
The given equation can be re-written as
[tex] y-18=14x^2-3x\\
\\
y-18=14(x^2-\frac{3}{14}x) \\
\\
y-18=14(x-\frac{3}{28})^2 -\frac{9}{56}\\
\\
y-18+\frac{9}{56}=14(x-\frac{3}{28})^2\\
\\
y-\frac{999}{56}= 14(x-\frac{3}{28})^2\\
\\
(x-\frac{3}{28})^2=\frac{1}{14}(y-\frac{999}{56})\\ [/tex]
As we know in standard form of [tex] (x - h)^2 = 4p (y - k) [/tex], the focus is (h, k + p)
So on comparision we get
[tex] 4p=\frac{1}{14}\\
\\
P=\frac{1}{56}\\ [/tex]
[tex] Focus=(\frac{3}{28},\frac{999}{56}+\frac{1}{56} )\\
\\
Focus=(\frac{3}{28}, \frac{125}{7}) [/tex]
