Find the Properties ((y+2)^2)/16-((x-3)^2)/9=1

Problem

((y+2)^2)/16−((x−3)^2)/9=1

Solution

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

2. Determine the center (h,k) From the terms (x−3) and (y+2) the center is (3,−2)

3. Calculate the semi-axes lengths. We have a^2=16 so a=4 (semi-transverse axis), and b^2=9 so b=3 (semi-conjugate axis).

4. Find the distance to the foci c using the relation c^2=a^2+b^2 Thus, c^2=16+9=25 which gives c=5

5. Locate the vertices by moving a units vertically from the center: (3,−2±4) This results in (3,2) and (3,−6)

6. Locate the foci by moving c units vertically from the center: (3,−2±5) This results in (3,3) and (3,−7)

7. Determine the equations of the asymptotes using the formula y−k=±a/b*(x−h) Substituting the values gives y+2=±4/3*(x−3)

Final Answer

Center: *(3,−2), Vertices: *(3,2),(3,−6), Foci: *(3,3),(3,−7), Asymptotes: *y+2=±4/3*(x−3)

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