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Graph ((y-3)^2)/16-((x+1)^2)/4=1

Problem

((y−3)2)/16−((x+1)2)/4=1

Solution

  1. Identify the type of conic section. Since the equation is in the form ((y−k)2)/(a2)−((x−h)2)/(b2)=1 it represents a vertical hyperbola.

  2. Determine the center (h,k) By comparing the given equation to the standard form, we find h=−1 and k=3 so the center is (−1,3)

  3. Calculate the values of a and b We have a2=16 which means a=4 and b2=4 which means b=2

  4. Locate the vertices. Since the hyperbola opens vertically, the vertices are at (h,k±a) which are (−1,3+4)=(−1,7) and (−1,3−4)=(−1,−1)

  5. Find the equations of the asymptotes. The asymptotes for a vertical hyperbola are given by y−k=±a/b*(x−h) Substituting the values, we get y−3=±4/2*(x+1) which simplifies to y−3=±2*(x+1)

  6. Determine the foci. Use the relation c2=a2+b2 to find c2=16+4=20 so c=√(,20)=2√(,5) The foci are at (h,k±c) which are (−1,3±2√(,5))

Final Answer

((y−3)2)/16−((x+1)2)/4=1⇒Vertical hyperbola centered at *(−1,3)* with vertices *(−1,7),(−1,−1)* and asymptotes *y=±2*(x+1)+3

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