The equation of a parabola is given. y=18x2+4x+20 What are the coordinates of the focus of the parabola?

The equation of a parabola is given. y=[tex] 18x^2+4x+20 [/tex]Here the value of 'a' = 18 and it is positive, it means the parabola opens up.For vertical parabola the focus point is (h , k+p)Where (h,k) is the vertex and p = [tex] \frac{1}{4a} [/tex] (p is the distance between vertex and focus)First we find vertex (h,k)h = [tex] \frac{-b}{2a} [/tex]a = 18 and b = 4So h= [tex] \frac{-4}{2*18} [/tex] = [tex] \frac{-4}{36} [/tex] = [tex] \frac{-1}{9} [/tex]Plug in -1/9 for x and find out kk = [tex] y = 18x^2+4x+20 = 18(\frac{-1}{9})^2 + 4(\frac{-1}{9}) + 20 [/tex]= [tex] (\frac{2}{9}) + (\frac{-4}{9}) + 20 [/tex]= [tex] \frac{178}{9} [/tex]So vertex is ([tex] \frac{-1}{9} [/tex], [tex] \frac{178}{9} [/tex])P = [tex] \frac{1}{4a} [/tex], we know a= 18So p = [tex] \frac{1}{4*18} [/tex] = [tex] \frac{1}{72} [/tex]Focus is ( h, k +p)h = [tex] \frac{-1}{9} [/tex] k = [tex] \frac{178}{9} [/tex]and p = [tex] \frac{1}{72} [/tex]k + P = [tex] \frac{178}{9} [/tex] + [tex] \frac{1}{72} [/tex] Take common denominator 72k + p = [tex] \frac{1424}{72} [/tex] + [tex] \frac{1}{72} [/tex] = [tex] \frac{1425}{72} [/tex] = [tex] \frac{475}{24} [/tex]Focus point is ( [tex] \frac{-1}{9} [/tex] , [tex] \frac{475}{24} [/tex])