how to find the focus of y =(1/8)x^2 -x +3
Well RULES FIRST: lets say that equation is x=y^2 + 4, in this one i know that focus is (6,0) but how do you obtain it? 1) Converting this into standard form, it is, "y^2 = x - 4" 2) A standard form of parabola equaiton is "(y - k)^2 = 4a(x - h)' whose, vertex is (h. k) and focus is (a + h, k) 3) Comparing the equation in (1) with the standard form in (2), h = 4, k = 0 and a = 1/4 4) Hence Focus is (4 + 1/4, 0) = (17/4, 0) You may kindly verify how it is (6, 0)
\[ y=\frac{x^2}{8}-x+3=\frac{1}{8} \left(x^2-8 x\right)+3=\\\frac {1}{8} \left((x-4)^2-16\right)+3=\frac{1}{8} (x-4)^2+1 \]
\[ y-1 =\frac{1}{8} (x-4)^2 \]
thanks so much!
Sorry Focus is (4,1+2)=(4,3)
See the attached graph.
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