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What is the focus of the parabola y=−10x2−3y = -10x^2 - 3?\newlineSimplify any fractions.\newline(______,______)

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Q. What is the focus of the parabola y=−10x2−3y = -10x^2 - 3?\newlineSimplify any fractions.\newline(______,______)
  1. Identify vertex form: Identify the vertex form of the parabola.\newliney=a(x−h)2+ky = a(x - h)^2 + k\newliney=−10x2−3y = -10x^2 - 3 can be rewritten as y=−10(x−0)2−3y = -10(x - 0)^2 - 3\newlineSo, h=0h = 0 and k=−3k = -3.
  2. Find value of 'p': Find the value of 'p' using the formula p=14ap = \frac{1}{4a}.p=14(−10)p = \frac{1}{4(-10)}p=−140p = -\frac{1}{40}
  3. Determine focus: Determine the focus of the parabola.\newlineThe focus is (h,k+p)(h, k + p).\newlineSubstitute h=0h = 0, k=−3k = -3, and p=−140p = -\frac{1}{40}.\newlineFocus = (0,−3−140)(0, -3 - \frac{1}{40})\newlineFocus = (0,−12040−140)(0, -\frac{120}{40} - \frac{1}{40})\newlineFocus = (0,−12140)(0, -\frac{121}{40})

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