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What is the focus of the parabola y=−8x2+1y= -8x^2+1? Simplify any fractions.

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Q. What is the focus of the parabola y=−8x2+1y= -8x^2+1? Simplify any fractions.
  1. Identify Parabola Form: Identify the form of the parabola and its orientation.\newlineGiven equation: y=−8x2+1y = -8x^2 + 1.\newlineThis is a vertical parabola since it's in the form y=ax2+ky = ax^2 + k.
  2. Convert to Vertex Form: Convert the equation to vertex form and identify aa, hh, and kk. The equation y=−8x2+1y = -8x^2 + 1 is already in the form y=a(x−h)2+ky = a(x - h)^2 + k where a=−8a = -8, h=0h = 0, and k=1k = 1.
  3. Calculate Value of p: Calculate the value of p using the formula p=14ap = \frac{1}{4a}.\newlineSubstitute a=−8a = -8 into p=14ap = \frac{1}{4a}.\newlinep=14(−8)=1−32=−132p = \frac{1}{4(-8)} = \frac{1}{-32} = -\frac{1}{32}.
  4. Determine Focus: Determine the focus of the parabola using the vertex (h,k)(h, k) and the value of pp. The vertex is (0,1)(0, 1). Since the parabola opens downwards (a<0)(a < 0), the focus is at (0,k+p)(0, k + p). Focus = (0,1−132)=(0,3132)(0, 1 - \frac{1}{32}) = (0, \frac{31}{32}).

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